Hurewicz tests: separating and reducing analytic sets the conscious way

By

Tamas Matrai

Submitted to

Central European University Department of Mathematics and its Applications

Supervisor: Tamas Keleti

Budapest, Hungary

2005

2

Abstract

In this thesis we develop the theory of topological Hurewicz test pairs. The central idea of this method is to construct special Polish topologies in such a way that simply structured sets, e.g. sets of low Borel class, are either “almost empty” or of second category in the corresponding special topology. Through this approach, which can be considered as a generalization of the Baire Category Theorem for higher Borel classes, we are able to give a quantitative aspect, in Baire category sense, to certain results related to Hurewicz test sets. As an application of the theory, we prove some results related to transnite convergence of Borel functions, to the problem of nding simple generators of analytic ideals and to the problem of constructing Hurewicz test sets for generalized separation of analytic sets.

Contents

1 Classical results, motivating problems 5

2 Denitions and notation 11

2.1 Sequences and trees ...... 11

2.2 Rening topologies, basic open sets ...... 12

2.3 Hierarchies, separation, reduction ...... 14

3 The = 3 case 17

0 3.1 The 3 set of Lusin ...... 17 3.2 Generating ideals ...... 24

4 Topological Hurewicz test pairs 37

4.1 Distinguishing Borel classes ...... 37

4.2 Intersection criteria ...... 45

4.3 A moment of being concrete ...... 53

4.4 Constructible coverings ...... 56

5 Testing the dierence hierarchy 59

6 -convergent functions 69 I 3

4 CONTENTS

7 Generalized separation and reduction 75 7.1 Witnessing generalized separation ...... 76 7.2 Generalized reduction ...... 85

8 Concluding remarks 87

Chapter 1

Classical results, motivating problems

If an open set is nonempty it is of second category. This is probably the simplest for- mulation of the Baire Category Theorem. The purpose of our thesis is to show that this

theorem can be extended to the entire Borel hierarchy: for every 0 < < ω1 we dene a

0 ne topology such that if a set is just “a bit bigger than nonempty” then it is huge, i.e. it is of second category in our ne topology. And we present some typical Baire Category Theorem-like applications of this result: the intersection of countably many “fairly dense”

0 sets is residual in the ne topology, hence it is nonempty. We remark that there is a quite classical extension of the dichotomy expressed by the Baire Category Theorem for open sets. More than a half century ago Witold Hurewicz

proved the following theorem, also called Hurewicz-dichotomy, about sets failing to be F.

Theorem 1.1. (W. Hurewicz) Let X be a Polish space and A X be an analytic set. If A is not F , then there is a continuous injection of the Cantor set into X, ϕ: 2ω X → 5

6 CHAPTER 1. CLASSICAL RESULTS, MOTIVATING PROBLEMS

ω 1 ω ω such that 2 ϕ (A) is countable dense in 2 ; that is ϕ(2 ) A is a relatively closed \ ∩ subset of A homeomorphic to the irrationals.

If we add the well-known fact that any subset N of the Cantor set 2ω homeomorphic

ω to the irrationals is never F we gave all the reasons why the pair (2 , N) is called the

Hurewicz test for F sets. Theorem 1.1 has been strengthened in many successive steps. In some sense the most general existence theorem for Hurewicz tests is the following (see [7]).

Theorem 1.2. (A. Louveau, J. Saint Raymond) Let 3 < ω and (X, ) be a Polish 1 ω 0 0 space. If P 2 is ( ω ) but not ( ω ) and A , A X is any pair of disjoint analytic 2 2 0 1 0 sets, then either A0 can be separated from A1 by a () set or there is a continuous one- ω ω to-one map ϕ: (2 , ω ) X with ϕ(P ) A and ϕ(2 P ) A . 2 → 0 \ 1 The same conclusion holds for = 2 if P 2ω is the complement of a dense countable 2 set.

0 0 0 Thus we may call a (2ω ) but not (2ω ) set P a Hurewicz test for (2ω ) sets, 0 in the sense that either a set A X is ( ω ), a fact which can be witnessed by a 2 0 description of the construction of A from simpler sets, or A is not (2ω ), a property which cannot be veried by checking some decomposition, but by Theorem 1.2 it can be

ω 1 witnessed by a continuous injection ϕ: 2 X satisfying ϕ (A) = P . → Theorem 1.2 turned out to be the ideal starting point of our investigations. It allows us

ω to isolate a simple reason, namely the image of the Hurewicz test ϕ(P) in ϕ(2 ) why the

0 analytic set A0 cannot be separated from the other analytic set A1 with one single set. But the aspect of dichotomy which was present in the Baire Category Theorem for open sets is lacking, in particular Theorem 1.2 says nothing about how “crude” a separation

0 attempt is if one insists on sets, or in other words, how well our A0 can be approximated

7

by 0 sets laying outside A . It was the problem of -convergent functions, raised by T. 1 I Natkaniec in [17], which shed light on this failure. We start with the denition.

Denition 1.3. Let be a cardinal, (X, ) and (Y, d) be Polish spaces. Consider an ideal on . We say that a sequence of functions f : X Y ( < ) -converges to the I → I function f : X Y , in notation f f, if → →I

< : f (x) = f(x) { 6 } ∈ I

for every x X. ∈

The question of T. Natkaniec is whether -convergence keeps the Baire classes or not, I that is under which condition on and it is true that if f is Baire- ( < ω) and I f f the limit function f must be also Baire-. Since the Baire class of a function →I is in close relation with the Borel class of its sublevel set it is not surprising that the problem of -convergent functions is equivalent to nding some quantitative analogue of I Theorem 1.2. It was the rst motivation for our investigations. In some sense a particular case of the problem of T. Natkaniec is the problem of generalized separation. It is well known that analytic sets have the generalized separation property, that is for every sequence (A ) of analytic sets with A = there is a i i<ω i<ω i ∅ sequence (B ) of Borel sets satisfying A B (i < ω) and TB = . But of what i i<ω i i i<ω i ∅ complexity must Bi (i < ω) be? That is, if is xed, when canTeach Bi (i < ω) be chosen

0 in ? This can be regarded as an approximation problem where the analytic sets Ai (i < ω) play the role of constraint. One can ask whether there is a theorem providing Hurewicz tests for generalized separation as Theorem 1.2 does for ordinary separation. The last problem we mention here has been posed by A. Miller. S. Solecki in [21] proved the following result.

8 CHAPTER 1. CLASSICAL RESULTS, MOTIVATING PROBLEMS

Theorem 1.4. (S. Solecki) Given a family of closed sets and an analytic set A in some I Polish space X, either A can be covered by the union of countably many members of or I A contains a nonempty 0 set G with the property that F G is meager in G for every 2 ∩ F . ∈ I Observe that this result has the feature of testing: if A can be covered by the union of countably many members of this is witnessed by the countable collection of closed I sets in realizing the covering, while the contrary, which is a priory unwitnessable, can I 0 be veried by a good 2 subset of A and Baire category. It is natural to ask whether this holds for every Borel class (see [15]).

Question 1.5. (A. Miller) For 2 < ω let be a -ideal which is generated by its 1 I 0 members. Is it true that for every analytic set A X, either A or there is a 0 ∈ I +1 set B A such that B / ? ∈ I These problems can be treated within the theory of topological Hurewicz test pairs, a concept we aim to develop in this thesis. Informally, it can be summarized as “sets of low Borel class are so simply structured comparing to sets of high Borel class that even Baire category can distinguish them in a suitable topology”. As an illustration, the precise topologized version of Theorem 1.2 is the following result (see [10], Theorem 4 on page 159).

0 Theorem 1.6. Let (X, ) be a Polish space. For every 2 < ω , there exist a ( ω ) 1 2 ω ω set P 2 and a Polish topology on 2 which is ner than ω such that P is nowhere P 2 dense and closed in the topology , and if an analytic set A X is P

0 ω 1. in (), then whenever for a continuous one-to-one mapping ϕ: (2 , ω ) (X, ) 2 → 1 we have that ϕ (A) P is of second category in P in the relative topology ∩ P |P 1 ω then ϕ (A) 2 is of second category in the topology ; P

9

0 ω 2. not in (), then there is a continuous one-to-one mapping ϕ: (2 , ω ) (X, ) 2 → 1 ω such that ϕ(P ) A and ϕ (A) 2 is of rst category in the topology . P

0 Moreover, if < 2ℵ is a cardinal and in our model the union of a number of meager sets is meager in Polish spaces the rst statement holds for every (not necessarily Borel)

0 set A which can be obtained as a union of a number of () sets.

This result extends Theorem 1.2 by providing the quantitative aspect we have been

looking for: the Baire category in the topology P . We remark that S. Solecki proved a

0 0 theorem where Baire category was used to distinguish sets from a given set (see

[22], Theorem 2.2). In his approach the topology P remains hidden. In the next chapters we will build up the theory of topological Hurewicz test pairs. The basic denitions and notations in Chapter 2 are followed by an introductory chapter, Chapter 3, where we make the reader familiar with the techniques and concepts in the = 3 case. In particular we show that the answer to a special form of the question of

A. Miller is consistently negative. In Chapter 4 we continue with the general 0 < < ω1

0 0 case, we prove Theorem 1.6 in an extended form and we construct coverings for generated ideals. Given that Theorem 1.2 can essentially be extended to the dierence hierarchy, we perform a partial extension for Theorem 1.6 in Chapter 5. Then in Chapter 6 we show that indeed, -convergence keeps the Baire class of functions. In Chapter 7 we I construct a test allowing the determination of the minimal complexity of sets involved in a generalized separation or reduction, and we conclude the thesis with Chapter 8 where we formulate some problems which hopefully will be as motivating as the ones presented here.1

Acknowledgements. The rst results of the author in the domain of Hurewicz test were obtained during a research stay in the Equipe d’Analyse Fonctionnelle de l’Universite

1The interested reader is referred to [11], [12], [13] and [14] for an improved treatment of this material.

10 CHAPTER 1. CLASSICAL RESULTS, MOTIVATING PROBLEMS

Pierre et Marie Curie - Paris 6 with the support of the Analysis and Operators European Training Network. The author is grateful to the members of the l’Equipe, in particular to the members of the descriptive set theory research group, with special emphasis on Gabriel Debs, Alain Louveau and Jean Saint Raymond for the discussions that provided a lot of help. The author would also like to thank the numerous remarks and comments received from colleagues outside France: Marton Elekes, Petr Holicky, Slawomir Solecki and Miroslav Zeleny. In the end, I would like to thank the endurance of my supervisor, Tamas Keleti who made continual eorts to read the manuscript through.

Chapter 2

Denitions and notation

2.1 Sequences and trees

ω Our terminology and notation follow [4]. Let (C, C ) denote the Polish space 2 with its usual product topology. For two nite sequences s, t ω<ω, we write s t (s t, resp.) ∈ if t is an extension (a proper extension, resp.) of s. The length of s is denoted by s . If | | _ s = (s0s1 . . . sn 1) and i < ω, then s i stands for the sequence (s0s1 . . . sn 1i). If T ω<ω is a subtree and s ω<ω then T = t ω<ω : s_t T . The terminal ∈ s { ∈ ∈ } nodes of T are denoted by T(T ).

Let , ϑ (i < ω) be ordinals. We write ϑ if is successor and ϑ + 1 = (i < ω) i i → i or if is limit, ϑ ϑ (i j < ω) and sup ϑ = . i j i<ω i For every ordinal < ω we x once and for all a sequence (ϑ ) such that ϑ . 1 i i<ω i → To avoid complicated notations, we do not indicate the dependence of the sequence on , it will be always clear which pair of ordinal and sequence is considered.

The order topology on an ordinal is denoted by o.

11

12 CHAPTER 2. DEFINITIONS AND NOTATION 2.2 Rening topologies, basic open sets

In this note we will notoriously rene Polish topologies by turning countably many closed sets into open sets. We do this as described in [4], that is the open sets of the ancient topology together with their portion on the members of our collection of closed sets serve as a subbase of the new, ner topology. We will use that the topology obtained in this way is also Polish.

Denition 2.1. Let (X, ) be a Polish space, = P : i < ω be a countable collection P { i } of 0() sets. Then [ ] denotes that Polish topology rening where each P (i < ω) is 1 P i turned successively into an open set.

It is easy to see that the resulting ner topology [ ] is independent from the enumer- P ation of . This will be clear shortly when we x a base of [ ]. We also use the notation P P [ ] when the countable collection of not necessarily 0() sets can be enumerated on P 1 P such a way that P is 0([ P : i < n ]). n 1 { i }

Denition 2.2. If n (n < ω) is a Polish topology on some base set X then n<ω n

denotes the coarsest topology on X which renes each n (n < ω). W

The resulting topology is also Polish and we will shortly x a countable base for it. Before doing this we need a precise notion of basic open sets in our spaces.

Denition 2.3. Let (X , ) (i I) be Polish spaces; if a basis is xed in the spaces i i ∈ Gi (X , ) (i I), which are meant to be the basic open sets in (X , ), then the basic open i i ∈ i i

sets of ( i I Xi, i I i) are the open sets of the form ∈ ∈ Q Q G X , i i i J i I J Y∈ ∈Y\ where J I is nite and G for every i J. i ∈ Gi ∈

2.2. REFINING TOPOLOGIES, BASIC OPEN SETS 13

If the basic open sets are xed in the Polish space (X, ) and [ ] makes sense for G P a countable collection of subsets of X, then the basic open sets of [ ] are of the form P P G F0 Fn 1 or G with G , Fi (i < n); a basic [ ]-open set is said to be ∩ ∩ ∩ ∈ G ∈ P P proper if it is not -open.

If the basic open sets are xed for the Polish topologies (n < ω) then the basic Gn n open sets for are the sets of the form G where G (m < ω, n < n<ω n i

Observe that the basic open sets dened on this way form a basis of i I i, [ ] and ∈ P

n<ω n, respectively. From now on whenever a Polish space (X, ) appQears we assume Wthat a countable basis comprised of basic -open sets is xed; and this is done with respect to the convention of Denition 2.3 if applicable. We take X to be basic -open. Basic open sets are assumed to be regular open. In zero dimensional spaces we assume that

0 our basic -open sets are 1(); note that our procedure of renement results a zero dimensional space from a zero dimensional one with 0([ ]) basic [ ]-open sets. We 1 P P denote ambiguously our special collection of basic -open sets also by .

The interior (closure, resp.) of a set A (X, ) is denoted by int (A) (cl (A), resp.). We will never have to x a special compatible metric on our Polish spaces but we will

condition on the diameter of sets. In this case diam denotes the diameter in an arbitrary xed metric generating . We assume that diam (X) 1. 0 We recall that a 2() subset G of the Polish space (X, ) is itself a Polish space with

the restricted topology G (see e.g. [4], (3.11) Theorem). In particular, the notions related to category in the topology make sense relative to G.

We will have to return to the topologies on the coordinates in product spaces. If (X, ), (Y, ) are arbitrary topological spaces and ( , ) = (X Y, ), then we dene X S Pr ( ) = . The projection of product sets in product spaces is dened analogously. X S

14 CHAPTER 2. DEFINITIONS AND NOTATION

If G X and G Y , we say that the set of product form G = G G is X Y X Y X nontrivial on the X coordinate if G = X. X 6

2.3 Hierarchies, separation, reduction

0 0 th As usual, () (() resp.) (0 < < ω) stands for the multiplicative (additive 0 0 resp.) Borel class in the Polish space (X, ), starting with 1() = closed sets, 1() = open sets. The dual class of a class 2X is dened by = A X : X A . { \ ∈ } 0 0 0 A set is called proper () if it is () but not () (0 < < ω1). We derive the dierence hierarchy as follows. Every ordinal can be uniquely written as + n where is limit and n < ω. We call even (odd, resp.) if n is even (odd, resp.).

Denition 2.4. Let 0 < ϑ, < ω1 and let (A)<ϑ be a sequence of subsets of a set X,

such that A A ( 0 < ϑ). Then D ((A ) ) X is dened by 0 ϑ <ϑ

x D ((A ) ) x A and the least < ϑ with x A ∈ ϑ <ϑ ⇐⇒ ∈ ∈ [<ϑ has parity opposite to that of ϑ.

With this operation in the Polish space (X, ) we set

0 0 D () = D ((A ) ): A (), A A ( 0 < ϑ) . ϑ ϑ <ϑ ∈ 0 0 0 For xed 0 < < ω1, a set is called proper Dϑ () if it is Dϑ () but not 0 Dϑ () (0 < ϑ < ω1). Next we give two denition related to ideals.

Denition 2.5. Let 0 < < ω . If (X, ) is a Polish space and P X, 0(P ) ( 0(P ), 1 S P 0 0 resp.) denotes the collection of () ((), resp.) subsets of P .

2.3. HIERARCHIES, SEPARATION, REDUCTION 15

Denition 2.6. Let be a -ideal and . We say that is generated by if for I F I I F every G there exists F (i < ω) such that G F . We say that is strongly ∈ I i ∈ F i<ω i I generated by if for every G there is an F sucSh that G F . F ∈ I ∈ F

We recall some denitions related to separation.

Denition 2.7. Let X be an arbitrary set. For A , A X we say that G X separates 0 1 A from A if we have A G X A . 0 1 0 \ 1 A class 2X has the separation property if for every A , A there is a set G 0 1 ∈ ∈ such that X G and G separates A from A . A class 2X has the generalized \ ∈ 0 1 separation property if for every sequence (A ) with A = there is a sequence i i<ω i<ω i ∅ (G ) such that (X G ) , A G (i < ω) andT G = . i i<ω \ i i<ω i i i<ω i ∅ X A class 2 has the reduction property if for every A,TA0 there are B, B0 ∈ ∈ such that B A, B0 A0, A A0 = B B0 and B B0 = . ∪ ∪ ∩ ∅ Our 2X has the generalized reduction property if for every sequence (A ) i i<ω there is a sequence (G ) such that G A , G G = (i, j < ω, i = j) and i i<ω i i i ∩ j ∅ 6

i<ω Gi = i<ω Ai. S S

16 CHAPTER 2. DEFINITIONS AND NOTATION

Chapter 3

The = 3 case

As we announced in the introduction, we start our work by constructing topological

0 Hurewicz test pairs for = 3 and then we prove the corresponding theorems for 2 generated ideals. From now on (X, ) denotes a Polish space.

0 3.1 The 3 set of Lusin

We will investigate the following object.

Denition 3.1. Let P X. We call the pair P, a 0() topological Hurewicz test { P } 3 pair in X if

0 1. P is a 3() set;

2. P is a Polish topology rening ;

0 3. P is a nowhere P -dense 1(P ) set;

4. for every 0() set A X and basic -open set G with G P = , if A P is 2 P ∩ 6 ∅ ∩ -residual in G P then A is -residual in G. P |P ∩ P

17

18 CHAPTER 3. THE = 3 CASE

0 As we shall see later, there exist 3() topological Hurewicz test pairs in uncountable Polish spaces. We present a construction providing such sets in a suciently large variety for our purposes. First we dene the topologies corresponding to P .

0 Denition 3.2. Consider a 3() set P and x a presentation

i i i P = P , P = Pj i<ω j<ω \ [ with 0() sets P i X (i, j < ω). Set 1 j

= P i : i < n, j < ω (n ω), = X P i : i < ω . Pn { j } P { \ }

We dene < = [ ] and = <[ ]. P Pω P P P Claim 3.3. We have the following.

0 < 0 1. P is 2(P ) and 1(P ).

2. If G is basic -open and G P = then G is in fact basic <-open. P ∩ 6 ∅ P

3. The topologies and < coincide. P |P P |P

4. If G is basic <-open and G X P i = (n i < ω) then G is basic [ ]-open. P ∩ \ 6 ∅ Pn

5. If G is a [ ]-open set then the topologies Pn

P i j and [ n] i j G ( i

Proof. The rst statement is obvious. Since = <[ ], proper basic -open sets P P P P do not intersect P , which shows 2. Form this we immediately get 3. By denition, if a set is basic <-open then it is basic [ ]-open for some n < ω. But P Pn i a basic [ ]-open but not [ ]-open set G0 satises G0 P (i < ω), so 4 follows. Pi+1 Pi

0 3.1. THE 3 SET OF LUSIN 19

i j For 5, it is enough to consider the case when G i

H G P i P j = H G P i P j ∩ ∩ \ 0 ∩ ∩ \ i

Theorem 3.4. With the notation of Denition 3.2, suppose that P j P i (i j < ω) and P i is [ ]-dense and [ ]-meager in X (i < ω); then Pi Pi

< 1. P is P -residual;

2. P, is a 0() topological Hurewicz test pair. { P } 3

i < < Proof. For 1, it is enough to show that P is P -dense (i < ω); being P -open this implies that each P i and hence P is <-residual. We have that P i is [ ] dense and P Pi contains P j (i j < ω). So if G is a basic <-open set then either G is [ ]-open and P Pi so P i G = , or G is not [ ]-open hence G P i. This proves the statement. ∩ 6 ∅ Pi For 2 we check the conditions of Denition 3.1: 1 holds by the choice of P . By

Denition 3.2, the topology P is Polish and renes , which proves 2.

0 For 3, by Claim 3.3.1 P is 1(P ) so it remains to show that P does not contain any nonempty basic -open set. Suppose that G P and G is nonempty basic -open. P P

20 CHAPTER 3. THE = 3 CASE

Then by Claim 3.3.2, G is basic <-open hence basic [ ]-open for some n < ω. But P n P Pn is [ ]-meager, so G P n which contradicts G P P n. Pn 6 For 4, let A X be 0(), G be a basic -open set with G P = and suppose 2 P ∩ 6 ∅ that A P is -residual in G P . Observe that since A is 0(), (0([ ]) (n ω), ∩ P |P ∩ 2 2 Pn 0 2(P ), resp.), the notions “meager” and “nowhere dense” coincide for A in the topology , ([ ] (n ω), , resp.). Pn P < < By Claim 3.3.2, G is actually P -open. Since by 1, P is P -residual in X and by Claim 3.3.3 the topologies and < coincide, we have that A is <-residual in G. Suppose P |P P |P P that A is not -residual in G; that is we have a nonempty basic -open set G0 G P P such that A G0 is -meager thus -nowhere dense in G0. So by passing to a nonempty ∩ P P j i basic -open subset we can assume that A G0 = . Since P P (i j < ω), P ∩ ∅ n < we have G0 = G00 (X P ) for some n < ω where G00 G is basic -open. Since ∩ \ P n G00 (X P ) = , from Claim 3.3.4 we get that G00 is basic [ ]-open. Now G00 G ∩ \ 6 ∅ Pn < < < implies that A is -residual hence -dense in G00. Since [ ] is coarser than , A is P P Pn P n [ ]-dense so [ ]-residual in G00. Thus A and X P are both [ ]-residual in G00 Pn Pn \ Pn n which yields A G00 (X P ) = , a contradiction which completes the proof. ∩ ∩ \ 6 ∅ The following theorem describes important additional properties of the construction in Theorem 3.4, and in fact the rst four statements can be considered as a reformulation. The fth statement points out an obvious fact.

Corollary 3.5. Let P, be a 0() topological Hurewicz test pair as in Theorem 3.4. { P } 3 Then the following hold.

1. If A X is 0() and <-residual then A is -residual. 2 P P

2. If A X is 0() and of <-second category then A is of -second category. 3 P P

3. If A X is 0() and of -second category then A is of <-second category. 2 P P

0 3.1. THE 3 SET OF LUSIN 21

4. If A X is 0() and -residual then A is <-residual. 3 P P

0 5. P is a proper 3() set.

0 < < 0 < Proof. Let A be 2() and P -residual. By Theorem 3.4.1, P is a P -residual 2(P ) set so A P = and A P is < -residual in P . By Claim 3.3.3 the topologies < and ∩ 6 ∅ ∩ P |P P |P coincide, so we have that A P is -residual in P . By Theorem 3.4.2, Denition P |P ∩ P |P 3.1.4 applies with G = X and we conclude that A is P -residual in X, as required.

0 For 2, let A = i<ω Ai with 2() set Ai (i < ω). As in the proof of 1, if A is of <-second categorySthen there exists an i < ω such that A P is < -residual in G P P i ∩ P |P ∩ for some basic <-open set G with G P = . Since by Claim 3.3.3 the topologies < P ∩ 6 ∅ P |P and coincide, we have that A P is -residual in G P . Thus G is a basic P |P i ∩ P |P ∩ -open set with G P = , by Theorem 3.4.2, Denition 3.1.4 applies and gives that A P ∩ 6 ∅ i is P -residual in G. Thus A is of P -second category, as required. Statements 3 and 4 follow from 1 and 2 by taking complements.

0 < Finally suppose that P is 3(). Since P is P -residual by Theorem 3.4.1, from 2 we

get that P is of P -second category. But P is P -nowhere dense by Denition 3.1.3. This

contradiction completes the proof.

We prove here another, a bit more technical lemma but of the same avor as Corollary 3.5.

Lemma 3.6. Let P, be a 0() topological Hurewicz test pair as in Theorem 3.4. { P } 3 0 Let A X be a P -meager 2() set. If G is a basic [ n]-open but not [ n 1]-open set P P for some n < ω, put

X,P ZG (A) = cl[ n](A G). P ∩

Then ZX,P (A) is [ ]-meager in G. G Pn

22 CHAPTER 3. THE = 3 CASE

Proof. Suppose that ZX,P (A) is of [ ]-second category in G. Now the 0() set P n G Pn 2 is [ ]-meager in G, so we have that A is of [ ]-second category in G P n. Since G is not Pn Pn \ j i n i j [ n 1]-open and P P (i j < ω), we have that G P = G i

Theorem 1.2 says in particular that in (C, C ) from the point of view of Hurewicz tests

0 each proper 3(C ) set is the same, up to passing to a perfect subset of C. Precisely we have the following.

0 Corollary 3.7. Let P, P 0 C be proper ( ) sets. Then there are continuous one-to- 3 C 1 1 one maps ϕ, ϕ0 : C C such that ϕ (P ) = P 0 and ϕ0 (P 0) = P . → 0 Theorem 3.4 shows that a fairly big family of 3(C ) sets P form a topological Hurewicz

test with the natural renement P of but in consequence the reasoning is at least

0 notationally complicated. So in view of Corollary 3.7 we can x one special 3(C ) set < for which the topologies P and P become simple without loosing to much generality. Since the method we apply had already been used by Lusin to build (probably the rst)

0 proper 3(C ) set, we denote it by PL (see also [9]). We carry out the construction in an arbitrary perfect Polish space (X, ). For every nite sequence s ω<ω, x a nonempty perfect set P X with the following ∈ s properties:

(3.1) P = X; ∅ <ω (3.2) P _ P _ = (s ω , i < j < ω); s i ∩ s j ∅ ∈ (3.3) P P and P is -nowhere dense in (P , ) (t s ω<ω); s t s |Pt t |Pt ∈ <ω (3.4) P _ is -dense in (P , ) (s ω ). s i |Ps s |Ps ∈ i<ω [

0 3.1. THE 3 SET OF LUSIN 23

To have P _ P (i < ω), one simply has to take a countable dense subset D = s i s s d , d , . . . P and cover successively every d with a perfect set P _ which is nowhere { 1 2 } s i s i dense in (P , ) and disjoint from P _ (j < i). Then (3.1-3.4) obviously hold. Once s C |Ps s j this done, let

(3.5) = P : s ω<ω, s < n (n ω), Pn { s ∈ | | } n n P = Ps (n < ω), PL = P .

s ω<ω n<ω ∈[ \ s = n | | Observe that this notation is in accordance with Denition 3.2. From now on if we

take a PL in X then we assume that the above construction has been carried out, such that (3.1-3.5) hold. Observe also that the conditions of Theorem 3.4 are satised: P j P i (i j < ω) by (3.3), P i is [ ]-dense in X by (3.4) and [ ]-meager in X by (3.3). So Pi Pi P , is a topological Hurewicz test pair, in particular P is a proper 0() set by { L PL } L 3 Corollary 3.5.5. Finally we formulate a corollary of Lemma 3.6, which gets particularly simple for P , . We x a notation in advance. { L PL }

Denition 3.8. For every A X set ZX,PL (A) = cl (A P ) (s ω<ω) and s ∩ s ∈

X,P X,P s +1 Z L (A) = P Z L (A) P | | . L ∪ s \ s ω<ω ∈[ <ω < Corollary 3.9. The topologies Ps and [ s +1] Ps coincide (s ω ). A basic P -open | P| | | ∈ L set is of the form G P where G basic -open and s ω<ω. ∩ s ∈ Let A X be a -meager 0() set. Then PL 2

X,P 1. ZX,PL (A) = Z L (A) (s ω<ω); s Ps ∈

2. ZX,PL (A) is -nowhere dense in P (s ω<ω); s |Ps s ∈

3. X ZX,PL (A) is < -dense in X. \ PL

24 CHAPTER 3. THE = 3 CASE

Proof. The coincidence of the topologies and the structure of basic < -open sets PL follow from (3.2) and (3.3). This gives

X,PL X,PL <ω Zs (A) = cl (A Ps) = cl[ s +1](A Ps) = ZPs (A) (s ω ), ∩ P| | ∩ ∈ which is 1.

X,PL <ω By Lemma 3.6, ZP (A) is [ s +1] Ps -meager in Ps (s ω ), so since “meager” s P| | | ∈ 0 and “nowhere dense” coincide for 2() sets, 2 follows from 1. < For 3, let G be a basic -open set, say G = G0 Ps where G0 is basic -open and PL ∩ <ω X,P X,P s +1 s ω . By (3.2) and (3.3) we have P Z L (A) Z L (A) P | | , thus ∈ s ∩ s ∪ X,P X,P X,P s +1 G X Z L (A) = G0 P (P Z L (A)) G0 P (Z L (A) P | | ). ∩ \ ∩ s \ s ∩ ∩ s \ s ∪ X,P s +1 Since Z L (A) is -nowhere dense in P by 2, and P | | is -meager in P by (3.3), s |Ps s |Ps s X,P s +1 G0 P (Z L (A) P | | ) = , which completes the proof. ∩ s \ s ∪ 6 ∅

3.2 Generating ideals

The next step is to show that if (X, ) is a Polish space and P X is a Borel but not 0 0 3() set then the -ideal generated by the 3() subsets of P can be covered by a -ideal strongly generated by some 0() subsets of P such that P / . This is surprising I 3 ∈ I 0 because as we will see later there are 3(C ) sets contained by PL which can be covered only by -residual 0( ) subsets of P . We think that this is the best covering result PL |PL 3 C L 0 for 3-generated ideals which can be obtained in ZFC. We will return to this question in the last chapter.

Theorem 3.10. Let (X, ) be an uncountable Polish space and P X be a proper 0() 3 set. Then there is a mapping : 0(P ) 0(P ) such that A (A) and S3 → P3 P (Ai) = (A, Ai 0(P ) (i < ω)). \ 6 ∅ ∈ S3 i<ω [

3.2. GENERATING IDEALS 25

Proof. First we construct = L for (X, ) = (C, C ) and P = PL. For every A 0(P ) x a presentation A = A where A is 0( ) (j < ω). Set ∈ S3 L j<ω j j 2 C S (A) = cl (A P ) (s ω<ω), s C j ∩ s ∈ j< s [| |

n(A) = s(A) (n < ω), L(A) = n(A). s =n m<ω m n<ω | [| \ [ It is clear that (A) is 0( ); (A) P n P m (m n < ω) shows (A) P . L 3 C n L L Since A P P n implies A (A) for j < n < ω, we also have A (A). It L j n L remains to show that if Ai 0(P ) with its xed presentation Ai = Ai (i < ω) ∈ S3 L j<ω j then we can nd a point in P (Ai). We do this by constructingS inductively a L \ i<ω L sequence s ω<ω, s = n (n

26 CHAPTER 3. THE = 3 CASE

We obtained that there is a basic -open set G G which satises G P = and C n ∩ sn 6 ∅ i G Psn C i n n(A ). We can pass to a basic C -open subset Gn+1 clC (Gn+1) G ∩ \ such that G S P = ; then we have (3.8). By (3.4) we can nd an s s ω<ω, n+1 ∩ sn 6 ∅ n n+1 ∈ s = n + 1 with P G = , thus (3.6), (3.7) and (3.9) hold. This choice | n+1| sn+1 ∩ n+1 6 ∅ completes the inductive step and the proof of the special case.

If (X, ) and P are arbitrary, by Theorem 1.2 we can take a continuous one-to-one

1 0 map ϕ: (C, ) (X, ) such that ϕ (P ) = P . For A (P ) let C → L ∈ S3

1 (A) = (P ϕ(P )) ϕ( (ϕ (A))). \ L ∪ L

Since homeomorphism preserves the Borel class of sets, this denition makes sense and

fullls the requirements.

0 One may say that the construction in Theorem 3.10 is trivial if the 3(C ) subsets of P (which are all -meager by Corollary 3.5.2) could be covered by a -meager L PL |PL PL |PL 0 3(C ) subset of PL. Then a category argument would give that

P (Ai) = (A, Ai 0(P ) (i < ω)). L \ 6 ∅ ∈ S3 i<ω [ However, this is not the case. The construction in the following claim has already been used by S. Solecki in [21] to prove Theorem 1.4.

Claim 3.11. There is a 0( ) set A P such that if B is 0( ) and A B P 3 C L 3 C L then B is -residual in P . PL |PL L

Proof. In this proof B(x, ε) denotes the -open ball centered at x C with radius C ∈ ε > 0. Fix a v ω<ω and a basic -open set V satisfying V P = . We dene an ∈ \ {∅} C ∩ v 6 ∅ <ω <ω <ω injection ϕ: ω ω , a map j : ω ω and basic C -open sets (Ut)t ω<ω with the → → ∈

3.2. GENERATING IDEALS 27

following properties.

(3.10) U = V, ϕ( ) = v; ∅ ∅ (3.11) s t = ϕ(s) ϕ(t); ⇒ <ω (3.12) U cl (U ) U P _ (s, t ω , t s); s C s t \ ϕ(t) j(t) ∈ (3.13) U U = (s, t ω<ω, t = s ); s ∩ t ∅ ∈ | | | | s <ω (3.14) diam (U ) 2| | (s ω ); C s ∈ <ω (3.15) U P _ = (s ω ); s ∩ ϕ(s) j(s) 6 ∅ ∈ <ω (3.16) x U P _ ε > 0 i < ω (U _ B(x, ε)) (s ω ). ∀ ∈ s ∩ ϕ(s) j(s) ∀ ∃ s i ∈

Suppose that the construction is done. Let Av,V = n<ω s ω<ω, s =n Us. This set is ∈ | | 0 2(C ), we show that Av,V PL. By (3.13) we have Av,VT = S ωω n<ω U n . By (3.14), ∈ | (3.15) and (3.3) we have S T

ω U n = U n Pϕ( n) PL ( ω ) | | ∩ | ∈ ∈ n<ω n<ω \ \ so indeed A P . v,V L <ω Next we show that U P _ cl (U A ) (s ω ). Let x U P _ . s ∩ ϕ(s) j(s) C s ∩ v,V ∈ ∈ s ∩ ϕ(s) j(s) By (3.16) there is a sequence (i ) ω such that U _ B(x, 1/l). By (3.12) and l l<ω s il <ω (3.14) we have A U = (s ω ), in particular A U _ = (l < ω) so indeed v,V ∩ s 6 ∅ ∈ v,V ∩ s il 6 ∅ x cl (U A ). ∈ C s ∩ v,V 0 < Finally we show that every (C ) set H C for which Av,V H is of -second 2 PL category in V P . Let H = H where H is 0( ) (i < ω). By the Baire Category ∩ v i<ω i i 1 C Theorem in A , there is anSi < ω such that H A is of -second category in v,V i ∩ v,V C |Av,V A , say H contains a basic -open set G A where G is basic -open. By v,V i C |Av,V ∩ v,V C <ω (3.14) there is an s ω such that U G. By (3.15) we have U P _ = . As ∈ s s ∩ ϕ(s) j(s) 6 ∅ we have seen above,

U P _ cl (U A ) cl (G A ) H . s ∩ ϕ(s) j(s) C s ∩ v,V C ∩ v,V i

28 CHAPTER 3. THE = 3 CASE

< Since Pϕ(s)_j(s) Pv by (3.10) and (3.11), Hi and thus H is indeed of -second category PL in V P . ∩ v To have A, x an enumeration (v , V ): i < ω of the pairs (v, V ) where v ω<ω, V is { i i } ∈ basic -open and V P = . Set A = A . If B is a 0( ) set containing A, say C ∩ v 6 ∅ i<ω vi,Vi 3 C B = B with B 0( ) then as we have shown above, B (i < ω) is of < -second i<ω i i 2 C S i PL category in every nonempty < -open set. So B (i < ω) and hence B is < -residual. T PL i PL < < Now if B PL then B is P residual in PL, by Claim 3.3.3 the topologies P and PL | L PL | L coincide, so B is -residual in P , as stated. PL |PL PL |PL L So it remains to make the construction; we do this recursively. Set U = V , ϕ( ) = v ∅ ∅ and let j( ) < ω be such that V Pv_j( ) = ; such a j( ) exists by (3.4). So we have (3.10- ∅ ∩ ∅ 6 ∅ ∅ 3.15) for s = . Suppose that we already have U , ϕ(s) and j(s) for s ω<ω, s n such ∅ s ∈ | | that (3.11-3.15) hold for s , t n and (3.16) holds for s < n. For every s ω<ω with | | | | | | ∈ s = n, using (3.3), we can x a countable set D = d (i): i < ω U P P _ | | s { s } s ∩ ϕ(s) \ ϕ(s) j(s) with the property that cl (D) = D cl (U P _ ). Then we can nd a basic - C ∪ C s ∩ ϕ(s) j(s) C open neighborhood Us_i of ds(i) such that

U _ P = , cl (U _ ) U P _ , U _ U _ = (i, j < ω, i = j) s i ∩ ϕ(s) 6 ∅ C s i s \ ϕ(s) j(s) s i ∩ s j ∅ 6

n+1+i and diam (U _ ) 2 (i < ω). Then (3.12-3.14) hold for s n + 1 and (3.16) C s i | | _ holds for s n. Dene ϕ(s) ϕ(s i) to have P _ U _ = (i < ω); this is possible | | ϕ(s i) ∩ s i 6 ∅ by (3.4). Then (3.11) holds for s , t n + 1. Again by (3.4), we can have j(s_i) < ω | | | | such that P _ _ _ U _ = (i < ω). Then also (3.15) holds for s n + 1. This ϕ(s i) j(s i) ∩ s i 6 ∅ | | completes the recursive step and the proof.

0 By assuming the Continuum Hypothesis we can cover the 3(C ) subsets of PL by 0 2(C ) sets such that PL is not in the -ideal generated by the covering sets. Combining 0 Theorem 3.4 with the previous example we see that in this case the covering 2(C ) sets cannot be the subsets of P : a 0( ) subset of P should be -meager in P while L 2 C L PL |PL L

3.2. GENERATING IDEALS 29

we have constructed a 0( ) subset of P which can be covered only by -residual 3 C L PL |PL 0 0 3(C ) subsets of PL. For this covering by 2() sets, Theorem 1.2 is too “imprecise” 0 in the sense that it does not care about the presentation of () sets. We prove an alternative for = 3.

0 Lemma 3.12. Let (X, ) be a Polish space, (Ps)s ω<ω be a family of 1() subsets of ∈ X satisfying (3.1-3.4). Let , P n (n < ω) and P be as in (3.5). Let A be a xed Pn L -meager 0() set. If x? X ZX,PL (A) and U X is basic -open with x? U then PL 2 ∈ \ ∈ ? there is a -compact set F X such that x F U A and (F Ps)s ω<ω satises ∈ \ ∩ ∈ (3.1-3.4) in the Polish space (F, ). |F Proof. Let = G : n < ω be an enumeration of the sets of the form G P G { n } ∩ s <ω <ω ? r +1 (s ω , G basic -open). Let r ω be such that x P P | | . We dene ∈ ∈ ∈ r \ recursively a tree T ω<ω and = (V , X ): s T such that T ωn (n < ω) is nite, T { s s ∈ } ∩ _ V X (s T ) is a basic -open set, X = x _ : i < ω, s i T X P (s T ) is s ∈ s { s i ∈ } \ L ∈ n n a nite set, and with V = s T, s =n Vs, X = s T, s =n Xs (n < ω) the following hold: ∈ | | ∈ | | S S (3.17) x? Xn Xn+1 (n < ω); ∈ _ (3.18) cl (V _ ) V U (i < ω, s, s i T ); s i s ∈ (3.19) V V = (s, t T, s = t ); s ∩ t ∅ ∈ | | | | s (3.20) diam (V ) < 2| | (s T ); s ∈ _ (3.21) V _ X = x _ (i < ω, s i T ); s i ∩ s { s i} ∈ (3.22) cl (G ) Xn = implies G V n+1 = (n < ω); n ∩ ∅ n ∩ ∅

(3.23) if i r is maximal such that xs Pr then Vs cl Pr A = (s T ); | | ∈ |i ∩ |i ∩ ∅ ∈ <ω m+1 (3.24) if xs Pt for some s T, t ω then Xs Vs (Pt m P ) = (m < t ); ∈ ∈ ∈ ∩ ∩ | \ 6 ∅ | | <ω (3.25) if x P for some s T, t ω then X V P _ = . s ∈ t ∈ ∈ s ∩ s ∩ t i 6 ∅ i<ω [

30 CHAPTER 3. THE = 3 CASE

Suppose that the construction is done, we show that

n n F = V = cl ( X ) n<ω n<ω \ [ fullls the requirements; here equality follows from (3.18), (3.20) and (3.21). By (3.18), (3.19) and (3.20), F is compact, (3.17) and (3.18) imply that x? F U. ∈ We show that F A = . If x F then by (3.19) and (3.20) there is a unique ωω ∩ ∅ ∈ ∈

such that x = n<ω V n . Let i r be maximal for which x n Pr i for innitely many | | | | ∈ |

n < ω. By (3.21T ) we have limn x n = x so x Pr . Fix an n < ω such that i is →∞ | ∈ |i

maximal for which x n Pr . Then by (3.23), V n cl (Pr A) = , in particular | ∈ |i | ∩ |i ∩ ∅ x Pr A, as stated. ∈ |i \ Now we prove that (3.1-3.4) holds for (F Ps)s ω<ω in the Polish space (F, F ). Since ∩ ∈ | (3.1) and (3.2) are automatic we only have to check (3.3) and (3.4). First we show that F P = cl ( Xn P ). It is clear that F P cl ( Xn ∩ s n<ω ∩ s ∩ s n<ω ∩ P ). For the reverse containment itSis enough to show that whenever G F SP = for s ∩ ∩ s 6 ∅ some basic -open set G then G Xn P = for some n < ω; so let G F P = . By ∩ ∩ s 6 ∅ ∩ ∩ s 6 ∅ regularity we can nd another basic -open set G0 cl (G0) G such that G0 F P = . ∩ ∩ s 6 ∅ n+1 n Let G = G0 P . Since G V = , by (3.22) we have cl (G ) X = hence n ∩ s n ∩ 6 ∅ n ∩ 6 ∅ G Xn P = , which proves the statement. ∩ ∩ s 6 ∅ For (3.3), let s t and suppose that G F P = for some basic -open set G. We ∩ ∩ t 6 ∅ have to show that G F P P = . By the preceding, x G P for some u T , ∩ ∩ s \ t 6 ∅ u ∈ ∩ t ∈ so by (3.17), (3.18), (3.20) and (3.21), x = x V G for some u v T . Then by u v ∈ v ∈ m+1 m+1 (3.24) for m = s < t we get Xv Vv (Pt m P ) = . Since Pt P , Xv F | | | | ∩ ∩ | \ 6 ∅ and V G, we have G F (P P ) = , as required. v ∩ ∩ s \ t 6 ∅ For (3.4), suppose that G F P = for some basic -open set G and t ω<ω. We ∩ ∩ t 6 ∅ ∈ have to nd some i < ω such that G F P _ = . Let x G P ; as above, we have ∩ ∩ t i 6 ∅ u ∈ ∩ t a u v T with x = x V G. Then by (3.25), X V P _ = , so from ∈ u v ∈ v v ∩ v ∩ i<ω t i 6 ∅ S

3.2. GENERATING IDEALS 31

V G and X F we have G F P _ = and we are done. v v ∩ ∩ i<ω t i 6 ∅ We do the construction recursively Ssuch that after the N th step of the recursion (3.17-

N 1 ? 3.23) hold for n N while (3.24) and (3.25) hold for xs X . Put T , set x = x ∈ ∅ ∈ ∅ ? and let x V U be a basic -open set such that V cl (Pr A) = ; this choice ∈ ∅ ∅ ∩ ∩ ∅ ? X,PL ? is possible since x / Z (A), which means that x / cl (Pr A). So x , V meet the ∈ ∈ ∩ ∅ ∅ requirements.

N N 1 N Suppose that T ω is dened, we have X and V such that (3.17-3.23) hold for ∩ N 1 n N, (3.24) and (3.25) hold for x X . We extend T up to level N + 1 and dene s ∈ XN , V N+1 such that (3.17-3.23) hold for n N + 1, (3.24) and (3.25) hold for x XN s ∈ as follows. Let s T , s = N be arbitrary; since x / P , to satisfy (3.24) and (3.25) ∈ | | s ∈ L for xs we have to take only nitely many points. We show that we can pick each of these

points x in Vs such that

(3.26) if i r is maximal for which x Pr then x / cl (Pr A). | | ∈ |i ∈ |i ∩

Let x P for some t ω<ω. For (3.24) let m < t , and let i r be maximal s ∈ t ∈ | | | |

such that Pt m Pr . We distinguish two cases. If Pt m = Pr then by Corollary 3.9.2, | |i | |i cl (P A) is -nowhere dense in P so by (3.3) we can pick a point in V P t m Pt m t m s t m | ∩ | | | ∩ | \ m+1 (P cl (Pt m A)). If Pt m Pr then by (3.2) for every y Pt m , i is the maximal ∪ | ∩ | |i ∈ |

for which y Pr . Since xs Pt m , we have Vs cl (Pr A) = by (3.23). By (3.3) we ∈ |i ∈ | ∩ |i ∩ ∅ m+1 can pick a point x in Vs Pt m P , such that x / cl (Pr A) follows from x Vs. ∩ | \ ∈ |i ∩ ∈ t +1 t +1 For (3.25), if x P | | then x X shows (3.25). So suppose that x / P | | . s ∈ s ∈ s s ∈ By (3.4) we have a j < ω such that V P _ = . By Corollary 3.9.2, cl (P _ A) s ∩ t j 6 ∅ t j ∩

is _ -nowhere dense in P _ so we can pick a point x in V P _ cl (P _ A). |Pt j t j s ∩ t j \ t j ∩ _ Let i r be maximal for which x Pr . If r i = t j then we have x / cl (Pr A). | | ∈ |i | ∈ |i ∩ _ If r i = t j then Pt Pr . We show that for xs, as well, i is the maximal such that | 6 |i t +1 xs Pr . If Pt = Pr then this follows from x / P | | while if Pt Pr then we are ∈ |i |i ∈ |i

32 CHAPTER 3. THE = 3 CASE

done by the maximality of i for x. Thus by (3.23), Vs cl (Pr A) = so x Vs gives ∩ |i ∩ ∅ ∈ x / cl (Pr A). ∈ |i ∩ Index x and the new points with s_i (i < n ) for some n < ω, put s_i T (i < n ) s s s ∈ s and take pairwise disjoint basic -open sets x _ V _ cl (V _ ) V such that (3.19), s i ∈ s i s i s (3.20), (3.22) and (3.23) hold; observe that (3.23) can be satised by (3.26). Since (3.17),

(3.18), (3.21), (3.24) and (3.25) hold, this completes the recursive step and the proof.

0 0 0 Instead of covering 3() sets by 2() sets we dene 2() sets avoiding them. The 0 sequence of 2() sets constructed for this must be specially nested.

Denition 3.13. Let (X, ) be a Polish space and consider PL in X. Let 0 < < ω1. Set 0 = X , while for 0 < < , let = B : i < ω be a collection of pairwise disjoint B { } B { i } 0() subsets of X. Set B = B ( < ), B = B. Let [] = [ ϑ], 1 i<ω i < ϑ< B < < ϑ ϑ [] = [ ] and PS[] = P [ ] (T ). We say that (S ) is PL PL ϑ< B L L ϑ< B B < PL-nested if S S

1. B B ( < );

2. B is < []-dense in X ( < ); PL

3. Bi is []-compact ( < , i < ω);

<ω 4. (Ps B)s ω satises (3.1-3.4) in the Polish space (B, [] B ) for every if ∩ ∈ | is successor and for < if is limit.

Since a compact Polish topology on a base set has no nontrivial compact Polish rene- ment, Denition 3.13.3 says in particular that on B the topologies [] ( ) coincide. i In the sequel we use this property without further reference. Next we show that Denition 3.13.4 holds for limit = , as well.

3.2. GENERATING IDEALS 33

Lemma 3.14. Let ( ) be a P -nested sequence for some < ω and suppose that B < L 1 A B is 0([] ) and [] -meager. Then 2 |B PL |B 1. B is < []-residual in X; PL

2. if is a limit ordinal then (Ps B )s ω<ω satises (3.1-3.4) in the Polish space ∩ ∈ (B , [] ); |B

B ,PL B < 3. B Z ∩ (A) is [] B -dense in B ; \ PL |

4. P B = . L ∩ 6 ∅ Proof. By Denition 3.13.1 and 2, B ( < ) is < []-dense and < []-open ( < ). PL PL Since a < []-open but not < []-open set is contained in B ( < ), B ( < ) is also PL PL < []-dense. Hence B ( < ) and so B is < []-residual in X, which is 1. PL PL For 2, we have (3.1) and (3.2) automatically. To have (3.3), let t s and suppose that G P = for some basic [] -open set. That is, G = G0 B B for some basic -open ∩ s 6 ∅ |B ∩ n ∩ set G0, < and n < ω. But (3.3) holds in B , in particular U = B G0 P P = . n n ∩ ∩ t \ s 6 ∅ < < Now U is a []-open set, so given that B is []-residual by 1, we have U B = , PL PL ∩ 6 ∅ as required. To see (3.4) x an s ω<ω and a basic [] -open set G with G P = . As before, ∈ |B ∩ s 6 ∅ we have that G = G0 B B for some basic -open set G0, < and n < ω. Now ∩ n ∩ (3.4) holds in B hence B G0 P _ = , say U = B G0 P _ = for some n n ∩ ∩ i<ω s i 6 ∅ n ∩ ∩ s i 6 ∅ i < ω. Again, U is a < []-open set, so given that B is < []-residual by 1, we have PL S PL U B = , which proves the statement. ∩ 6 ∅ We have 3 since for limit by 2, for successor by Denition 3.13.4, Corollary 3.9.3 holds in the Polish space (B , [] ). |B Finally by 2 for limit and by Denition 3.13.4 for successor, Theorem 3.4.1 applies

< and we get that PL B is [] B -residual in B ; in particular, PL B = . This shows ∩ PL | ∩ 6 ∅

34 CHAPTER 3. THE = 3 CASE

4 and completes the proof.

Now we have everything to prove the ideal generation theorem. We remark that CH

0 is used only to assure that the cardinality of the family of 2() sets is ω1.

Theorem 3.15. Assume CH. Let (X, ) be a Polish space and P X be a Borel not 0() set. Then there is a -ideal such that 3 I

1. is strongly generated by its 0() members; I 2

2. 0(P ) ; S3 I

3. P / . ∈ I

Proof. First we prove the special case (X, ) = (C, C ), P = PL. Using CH, let A : < ω be an enumeration of 0(P ) such that A = . We shall construct a { 1} P2 L 0 ∅ PL-nested sequence ( ) such that B <ω1

(3.27) A B = ( < ω ) . ∩ ∅ 1

Once this done set = G C : < ω (B G = ) , IPL { ∃ 1 ∩ ∅ } or, since B B B ( < ω ) by Denition 3.13.1, equivalently +1 1

= G C : < ω (B G = ) . IPL { ∃ 1 ∩ ∅ }

By Denition 3.13, is a -ideal. Also by Denition 3.13.1, is strongly generated IPL IPL by its 0 ( ) members. By (3.27) it contains 0 (P ), hence 0 (P ), as well. Finally 2 C P2 L S3 L Lemma 3.14.4 implies that P / , as required. L ∈ I It remains to make the construction. We proceed by induction; to start with, set 0 = B0 with B0 = C. Suppose that is dened for < such that the sequence B { 0 } 0 B

3.2. GENERATING IDEALS 35

( ) is P -nested and (3.27) hold. By Denition 3.13.4 if is a successor and by B < L Lemma 3.14.2 if is limit, P B , [] is a topological Hurewicz test pair; in { L ∩ PL |B } particular, P B is [] meager in B . Since A B P B and , A B is L ∩ PL |B ∩ L ∩ ∩ a [] -meager 0([] ) set. So by Lemma 3.14.3 we can x a countable set PL |B 2 |B

B ,P B Y = x : n < ω B Z L∩ (A B ) { n } \ ∩

< < such that Y is [] B -dense in the Polish space (B , [] B ). We dene recursively PL | PL | B (n < ω) such that B = B (n < ω), ( ) is P -nested and (3.27) hold. n i

36 CHAPTER 3. THE = 3 CASE

0 For an arbitrary uncountable Polish space (X, ) and Borel not 3() set P take a 1 continuous one-to-one map ϕ: (C, ) (X, ) such that ϕ (P ) = P . Let C → L

= G X : G0 (G ϕ(C) ϕ(G0)) . I { ∃ ∈ IPL ∩ }

Since homeomorphism preserves the Borel class of sets and X ϕ(C) is -open hence \ 0(), this -ideal is strongly generated by its 0() members. If A P is 0() then 2 2 3 1 0 1 0 ϕ (A) P is ( ) so ϕ (A) and hence A . This shows (P ) . Since L 3 C ∈ IPL ∈ I S3 I P / we have P / , which completes the proof. L ∈ IPL ∈ I Let us turn back to Question 1.5. The analogous question for strong generation is the following.

Question 3.16. Let (X, ) be a Polish space and let 0 < < ω . If a -ideal 2X 1 I is strongly generated by its 0() members and a Borel set A X is not in then does I there exists a 0 () set B A such that B / ? +1 ∈ I

If = 1 then the answer is obviously armative, since if B is a countable -dense |A subset of A then B / is shown by A cl (B). ∈ I Theorem 3.15 shows that for = 2 the answer is consistently negative. We think that Question 3.16 is independent. In particular, we belive that Theorem 3.10 gives the best constructible covering for the ideal 0(P ). We will come back to this problem in Chapter S3 8.

Chapter 4

Topological Hurewicz test pairs

This chapter is devoted to the extensions of the results of Chapter 4 we have for 4 < 0 ω1: the existence of topological Hurewicz test pairs and the ideal generation theorem.

4.1 Distinguishing Borel classes

In this section we extend Theorem 3.4 to higher levels of the Borel hierarchy. In order to produce a suciently big family of test pairs we need a machinery which allows us to condition on the construction of a given Borel set from simpler sets. For this, we handle

0 a () set by coding its construction from closed sets in a tree. The following inductive denition makes this concrete.

0 Denition 4.1. Let 0 < < ω1 and ϑi . For = 1, [P, (P )] is called a 1() set → ∅ 0 with presentation if P = P is a 1() set. ∅ 0 Suppose that the ϑ() sets with presentation are dened for ϑ < . Then [P, (Pt)t T ] ∈ is a 0() set with presentation if T ω<ω is a subtree such that (i): i < ω T , { } 0 _ P = X P(i) and [P(i), (Pi t)t T(i) ] is a ϑ () set with presentation (i < ω). \ i<ω ∈ i S 37

38 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

0 It is important to note that a () sets with presentation is not necessarily a proper 0 () set. For example, it can easily be empty. Next we dene the test sets and the corresponding topologies. These test pairs are ex- actly what one may expect; in particular, the following denition for = 3 is in accordance with Denition 3.1.

Denition 4.2. Let 0 < < ω and P X. We call the pair P, a 0() topological 1 { P } Hurewicz test pair in X if

0 1. P is a () set;

2. P is a Polish topology rening ;

0 3. P is a P -nowhere dense 1(P ) set;

4. (a) = 1: for every -open set A X and basic -open set G with G P = if P ∩ 6 ∅ A P is -residual in G P then A is -residual in a -open set G0 G ∩ P |P ∩ P P such that G P cl (G0 P ). ∩ P ∩ (b) 1 < is a successor ordinal: for every ϑ < , 0 () set A X and basic ϑ -open set G with G P = if A P is -residual in G P then A is P ∩ 6 ∅ ∩ P |P ∩ P -residual in G.

(c) 1 < is a limit ordinal: there is a -open set (ϑ) (ϑ < ) such that P HX,P P (ϑ) (ϑ < ), and for every ϑ < , 0 () set A X and basic HX,P ϑ -open set G with G P = if A P is -residual in G P then A is P ∩ 6 ∅ ∩ P |P ∩ -residual in G (ϑ). P ∩ HX,P

< 0 We associate inductively the topologies P and P to () sets with presentation

(0 < < ω1).

4.1. DISTINGUISHING BOREL CLASSES 39

0 Denition 4.3. Consider a () set with presentation [P, (Pt)t T ]. For = 1 we dene ∈ < 0 P = P = . If 1 < < ω1 and P is dened for ϑ() sets with presentation for ϑ < , set = P (X P ): n < ω . We dene < = and = <[ ]. P { (n) ∩ i

Theorem 4.4. (Kuratowski-Ulam) Let (X, ) and (Y, ) be Polish spaces, let G = G X G be a basic -open set in X Y and consider a Borel set A X Y . Set Y Ay = x X : (x, y) A . Then A is -residual in G if and only if { ∈ ∈ }

y G : Ay is -residal in G { ∈ Y X }

is -residual in GY .

Claim 4.5. With the notation of Denition 4.3, we have the following.

0 < 0 1. P is 2(P ) and 1(P ).

2. If G is basic -open and G P = then G is in fact basic <-open. P ∩ 6 ∅ P 3. The topologies and < coincide. P |P P |P 4. The topologies

< P P P and P P P (n < ω) | (n)\ i

5. If (Y, ) is any nonempty Polish space and P, is a 0() topological Hurewicz { P } test pair in (X, ) then P Y, is a 0( ) topological Hurewicz test pair { P } in (X Y, ).

40 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

Proof. We prove the rst statement by induction on . For = 1 the statement is

obvious. Let now 1 < < ω1 and suppose that the statement holds for ϑ < . We have

(4.1) P = X P = X P (X P ) , \ (n) \ (n) ∩ \ (i) n<ω n<ω i

G0 = H X Y : H is basic -open, A is -residual in H . { P P } [ Then A is P -residual in G0 so it remains show that G (P Y ) cl (G0 (P ∩ P ∩ Y )). Suppose that K = K K G is a nonempty basic -open set such that X Y P K (P Y ) G (P Y ) cl (G0 (P Y )) . Then A (P Y ) is P -residual ∩ ∩ \ P ∩ ∩ in (K P ) K , so by Theorem 4.4, X ∩ Y W = y K : Ay P is -residual in K P { ∈ Y ∩ P |P X ∩ } is -residual in K . Since P, is a 0()-topological Hurewicz test pair, by Denition Y { P } 1 y 4.2.4a, for every y W there is a -open set K0 (y) K such that A is -residual in ∈ P X X P K0 (y) and K P cl (K0 (y) P ). Since (X, ) has countable base, K0 (y) contains X X ∩ X ∩ P X the same basic P -open set for -residually many y’s, that is there is a basic P -open set

K0 K and a basic -open set K0 K such that X X Y Y y y K0 : A is -residual in K0 { ∈ Y P X }

4.1. DISTINGUISHING BOREL CLASSES 41

is -residual in K0 . Then by Theorem 4.4, A is -residual in K0 = K0 K0 , that Y P X Y is K0 G0, a contradiction. Let now 1 < < ω be a successor ordinal, ϑ < and A be a 0 ( ) set such that 1 ϑ A (P Y ) is -residual in (G P ) G . We show that A is -residual in ∩ P X ∩ Y P G. By Theorem 4.4,

W = y G : Ay P is -residual in G P { ∈ Y ∩ P |P X ∩ } is -residual in G . Since P, is a 0()-topological Hurewicz test pair, by Denition Y { P } 4.2.4b, Ay (y W ) is -residual in G . Then again by Theorem 4.4, A is -residual ∈ P X P in G, as stated.

Let now 1 < < ω1 be a limit ordinal. We show that X Y,P Y (ϑ) = X,P (ϑ) Y H H fullls the requirements. Let ϑ < and A be a 0 ( ) set such that A (P Y ) is ϑ ∩ -residual in (G P ) G . By Theorem 4.4, P X ∩ Y W = y G : Ay P is -residual in G P { ∈ Y ∩ P |P X ∩ } is -residual in G . Since P, is a 0()-topological Hurewicz test pair, by Denition Y { P } 4.2.4c, Ay (y W ) is -residual in G (ϑ). Then again by Theorem 4.4, A is ∈ P X ∩ HX,P P -residual in G X Y,P Y (ϑ) = (GX X,P (ϑ)) GY , as stated. This completes ∩ H ∩ H the proof.

The following claim describes the behavior of a topological test pair with respect to

0 () sets.

Claim 4.6. Let 0 < < ω and let P, be a 0() topological Hurewicz test pair. If 1 { P } for a 0() set W and -open set G with G P = we have that W P is -residual P ∩ 6 ∅ ∩ P |P in G P , then W is -residual in a -open set H satisfying that G P cl (H P ). ∩ P P ∩ P ∩

Proof. For = 1 the statement follows from the denition. Let now 1 < < ω1 and

0 write W = Qi, where Qi is () and ϑi . If W G P is P P -residual in i<ω ϑi → ∩ ∩ | S

42 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

G P then let H denote the maximal -open set in which Q is -residual (i < ω). ∩ i P i P

By Denition 4.2.4, the P -open set H = i<ω Hi meets every P -open set intersecting G P , which proves the statement. S ∩ In the following theorem we give a method allowing to build up inductively a topo- logical Hurewicz test pair from simpler test sets .

0 Theorem 4.7. Let 0 < < ω1, ϑi and let [P, (Pt)t T ] be a nonempty () set with → ∈ presentation. If = 1 and P is -nowhere dense then P, is a topological Hurewicz { P } test pair.

< < 0 For 1 < < ω1 suppose that P is -dense and P , is a () topological i<ω (i) P { (i) P } ϑi Hurewicz test pair (i < ω). ThenS

< 1. P is P -residual;

2. P, is a 0() topological Hurewicz test pair. { P }

< Proof. If = 1 then G0 = A G does the job. Let now 1 < < ω . Since P , ∩ 1 { (n) P } 0 < (n < ω) is a ϑn () topological Hurewicz test pair, P(n) (n < ω) is P -nowhere dense so 1 follows from (4.1). For 2 we have to check the conditions of Denition 4.2; 1 holds by the choice of P , 2 follows from Denition 4.3. For 3, by Claim 4.5.1 it remains to show that P does not contain any nonempty basic -open set. Suppose that G P and G is nonempty basic -open. Then by Claim P P 4.5.2, G is basic <-open, we have that P is <-dense hence P G = for some P i<ω (i) P (n) ∩ 6 ∅ n < ω, a contradiction. S Let now ϑ < , A X be 0 (), G be a basic -open set with G P = and suppose ϑ P ∩ 6 ∅ that A P is -residual in G P . By Claim 4.5.2, G is actually <-open while by 1 ∩ P |P ∩ P < and Claim 4.5.3, A is P -residual in G.

4.1. DISTINGUISHING BOREL CLASSES 43

Set G0 = G if is a successor. If is limit then let I < ω be minimal such that ϑ ϑ , I set (ϑ) = X P and G0 = G (ϑ) = G P . We have P (ϑ) HX,P i

Suppose that A is not P -residual in G0; that is we have a nonempty basic P -open

set G˜ G0 such that A G˜ is -meager in G˜. By passing to a nonempty basic -open ∩ P P subset we can assume that

G˜ = G P X P = G P P 0 ∩ (n) ∩ \ (i) 0 ∩ (n) \ (i) i

Just as for the = 3 case, the conditions of Theorem 4.7 concern the presentation of

0 the () set P instead of P itself. This handicap seems to be inevitable. First, because 0 up to our knowledge there are no results providing some method to build up () sets 0 0 from simpler sets, there is not even a canonical decomposition of 2() sets to 1() sets.

It is easy to see that by taking a wrong presentation the topology P becomes wrong

0 either, that is we cannot just condition our test set to be proper (), a suitably chosen, not natural presentation must be involved. Second, because the only way to build up a

44 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

0 proper () set for > 4 is to use induction, so in view of our rst reason one could hardly imagine Theorem 4.7 without some inductive condition on the presentation. Even if well explained, this handicap remains painful and this is responsible for most of the complication we have to face later. We close this section with the usual corollaries.

Corollary 4.8. For a 0 < < ω , let P, be a 0() topological Hurewicz test pair 1 { P } as in Theorem 4.7. Let G be a basic -open set with G P = , or equivalently let G be P ∩ 6 ∅ < a nonempty basic P -open. Then the following hold.

1. If is a successor, ϑ < and A X is 0 () and <-residual in G then A is ϑ P P -residual in G.

2. If A X is 0() and of <-second category in G then A is of -second category P P in G.

3. If is a successor, ϑ < and A X is 0 () and of -second category in G then ϑ P < A is of P -second category in G.

4. If A X is 0() and -residual in G then A is <-residual in G. P P

0 5. P is a proper () set.

0 < < 0 < Proof. Let A be ϑ() and P -residual in G. Since P is a P -residual 2(P ) set, A P is < -residual in G P . Since by Claim 4.5.3 the topologies < and ∩ P |P ∩ P |P P |P coincide, A P is -residual in G P . So Denition 4.2.4b applies and we conclude ∩ P |P ∩ that A is P -residual in G, which proves 1.

0 For 2, let A = Ai with () set Ai (i < ω) where ϑi . Since P is a i<ω ϑi → <-residual 0( <) Sset, if A is of <-second category in G then for an i < ω, A P P 2 P P i ∩ < < is -residual in G0 P for some basic -open set G0 G. Since by Claim 4.5.3 P |P ∩ P the topologies < and coincide, we have that A P is -residual in the basic P |P P |P i ∩ P |P

4.2. INTERSECTION CRITERIA 45

-open set G0 with G0 P = so by Denition 4.2.4, A is -residual in some nonempty P ∩ 6 ∅ i P -open set G00 G0 thus A is of -second category in G, as required. P P Statements 3 and 4 follow from 1 and 2 by taking complements.

0 < For 5, suppose that P is (). By Theorem 4.7.1, P is P -residual in X so by

Corollary 4.8.2, P is of P -second category in X. But by Denition 4.2.3, P is P -nowhere

dense, a contradiction. This completes the proof.

There is an asymmetry in our approach to topological Hurewicz test sets: the test set is of some multiplicative class and the sets tested are of the dual additive class. The

0 0 reason for this is that is closed under taking countable union while is not. However, 0 there is a testing theorem like Theorem 4.7 for special sets (see e.g. Theorem 5.2 or Corollary 5.3) but the statement of this theorem cannot go beyond Corollary 4.8. So we do not work for that.

4.2 Intersection criteria

It turns out that the conditions of Theorem 4.7 are combinatorial, they are the same as requiring that countably many intersections are nonempty. We give these intersections. Our purpose is to show that if Theorem 4.7 proves that a set P is a topological Hurewicz test then P remains a test set if the initial topology of the Polish space is changed.

0 Denition 4.9. Let 0 < < ω1 and [P, (Pt)t T ] be a () set with presentation. If ∈

= 1, set (X, , P ) = (X P, G): G . If (X0, 0, P 0) is dened for every ϑ < , C1 { \ ∈ } Cϑ 0 Polish space (X0, 0) and () set with presentation P 0 then let ϑ and set ϑ i →

(X, , P ) = (X P, G): G < X, , P . C { \ ∈ P } ∪ Cϑi P(j) (i) i<ω j<ω, j=i ! [ _ 6

46 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

We say that [P, (Pt)t T ] satises in (X, ) if ∈ {

(C, G) (X, , P ) (G = = C G = ). ∀ ∈ C 6 ∅ ⇒ ∩ 6 ∅

0 Claim 4.10. Let 0 < < ω1. If a () set with presentation [P, (Pt)t T ] satises in ∈ { (X, ) then P, satises the conditions of Theorem 4.7, so in particular P, is a { P } { P } 0 () topological Hurewicz test pair.

Proof. We prove the statement by induction on . For = 1, {1 means that P is -nowhere dense in X, that is P, is a 0() topological Hurewicz test pair by { P } 1 Theorem 4.7. Suppose now that the statement holds for ϑ < and let ϑ . By , i → { < _ X P = i<ω P(i) is P -dense in X and [P(i), (Pi t)t T(i) ] satises {ϑi in the Polish space \ ∈

(X, j<ω,Sj=i P(j) ). We have 6 W < P = P(j) = P(j) , j<ω j<ω, j=i ! _ _ 6 P(i) so by the induction hypothesis P , < is a 0 () topological Hurewicz test pair. Thus { (i) P } ϑi the conditions of Theorem 4.7 are satised, Theorem 4.7.2 can be applied and we conclude that P, is a 0() topological Hurewicz test pair. { P } We need that if P lives in a product space but it is nontrivial only on one coordinate

then { also conditions only on one coordinate.

0 Claim 4.11. Let 0 < < ω1 and let [P, (Pt)t T ] be a () set with presentation in ∈ the Polish space (X, ). Let (Y, ) be a Polish space and set Q = P Y , Q = P Y t t 0 (t T ). Then [Q, (Qt)t T ] is a ( ) set with presentation and for every (C, G) ∈ ∈ ∈ (X Y, , Q), C is nontrivial only on the X coordinate. C

Proof. The statement easily follows by induction on .

The next claim gives that { remains true if the initial topology gets coarser.

4.2. INTERSECTION CRITERIA 47

0 Claim 4.12. Let 0 be a Polish topology on X rening . If for some 0 < < ω1 a ()

set with presentation [P, (Pt)t T ] satises in (X, 0) then it satises in (X, ) as well. ∈ { {

Proof. We prove by induction on that (X, , P ) (X, 0, P ) (0 < < ω ). C C 1 From this the statement follows.

For = 1 we have 0 and so (X, , P ) (X, 0, P ). Suppose now that the C1 C1 < < statement holds for ϑ < and let ϑi . Since 0 we also have 0 , P 0 → P P (i) P(i) (i < ω) so

< < (X P, G): G (X P, G): G 0 { \ ∈ P } { \ ∈ P } and by the induction hypothesis,

ϑ X, P , P(i) ϑ X, 0 , P(i) . C i (j) C i P(j) j<ω, j=i ! j<ω, j=i ! _ 6 _ 6 This proves (X, , P ) (X, 0, P ) and completes the proof. C C From now on in this section we work to prove the main lemma of the extension of Theorem 3.10. The technique of the proof is to exploit the low class Hurewicz test sets

0 appearing in the construction of a () test set. For this we need some more topologies.

0 Denition 4.13. Let 0 < < ω1 and let [P, (Pt)t T ] be a () set with presentation ∈ which satises { in (X, ). We dene the topologies

< P (n) = P (n < ω). (i) ∨ P(i) i n n

< < P (0) = Pi i<ω < _ and P (n + 1) = P (n) (n < ω).

48 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

_ Proof. Since [P, (Pt)t T ] satises , [P(n), (Pn t)t T ] satises ϑn in the Polish space ∈ { ∈ (n) {

_ (X, j<ω, j=i P(j) ). Then by Claim 4.12, [P(n), (Pn t)t T(n) ] satises {ϑn in the Polish space 6 ∈ W < X, P . (i) ∨ P(i) i

< P (n) = P P (n) ∨ (i) ∨ P(i) i

< < < P (n + 1) = P = P (n) (n < ω) P(n+1) ∨ (i) ∨ P(i) i

< < < < P (0) = = , P(0) ∨ P(i) P(i) 0

The next lemma is an application of our newly found Hurewicz test sets.

Lemma 4.15. Let 2 < < ω1 be such that we have = 0 + 1 where 0 = 00 + 1 is

0 a successor. Let [P, (Pt)t T ] be a () set with presentation which satises . Fix an ∈ { n < ω. If A is a 0 () set and A is (n)-meager in a (n)-open set G then A is also 00 P P (m)-meager in G (n m < ω). P

Proof. If G is (n)-open then it is (m)-open (n m < ω) so it is enough to prove P P that A is P (n + 1)-meager in G; from this the statement follows by induction.

0 By Corollary 4.14 we have that P(n+1), P (n+1) is a () topological Hurewicz test { } 0 < pair and P (n + 1) = P (n). So by Corollary 4.8.3 our A cannot be of P (n + 1)-second

category in G. This completes the proof.

4.2. INTERSECTION CRITERIA 49

0 The crucial point in the proof of Theorem 3.10 was the fact the a 2() set of rst category is nowhere dense whatever is the underlying Polish topology. This is not true for 0 () sets when 3 ϑ < ω so in order to avoid them we have to reduce complicated ϑ 1 0 sets to 2() sets. This is the motivation of the following concept.

<ω Denition 4.16. Let T ω be a tree. We say that a subtree T 0 T is even-complete if

_ t T 0, t odd = t t 1 i: i < ω T 0 = t & ∈ | | ⇒ || | ∩ { } _ _ t i: i < ω T = & t i: i < ω T T 0, { } ∩ 6 ∅ { } ∩

and T 0 is maximal with respect to this property.

+T If T 0 T is an even complete subtree, t T 0 T(T 0) with t even then t T 0 ∈ \ | | ∈ denotes the unique extension of t with t+T = t + 1. | | | |

0 Lemma 4.17. For some 0 < < ω1 let [P, (Pt)t T ] be a () set with presentation. If ∈

T 0 T is an even-complete subtree and x / P (t T(T 0)) then x / P . ∈ t ∈ ∈ Proof. We prove the statement by induction on . For = 1 and = 2 the only

even-complete subtree of T is T 0 = that is x / P = P , as stated. Suppose now that {∅} ∈ ∅ 3 and the statement holds for ϑ < . By maximality we have T 0 = , hence there is 6 ∅

a unique i < ω such that (i) T 0. But T 0_ is an even-complete subtree of T _ (j < ω) ∈ i j i j so by the induction hypothesis x / P _ (j < ω). That is x P and so x / P , which ∈ i j ∈ (i) ∈ completes the proof.

Now we can prove the main result of the section.

Claim 4.18. Let 1 < < ω1 be such that we have = 0 + 1 where 0 = 00 + 1 is a

0 n successor. Let [P, (Pt)t T ] be a () set with presentation which satises . If A is a ∈ { 0 n n () set such that A P(n) = (n < ω) then P A = . 0 ∩ ∅ \ n<ω 6 ∅ S

50 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

n n Proof. Fix a presentation [A , (At )t T n ] (n < ω). Take maps 1 : ω ω and 2 : ω ∈ → → ω<ω such that = ( , ): ω n T n 1 2 → { } n<ω [ is a bijection satisfying

n 1 1 (4.2) t, t0 T , t t0 = (n, t) (n, t0) (n < ω) ∈ ⇒

and

(4.3) (n) max 0, n 1 (n < ω). 1 { }

We construct inductively a basic P (n)-open set Gn (n < ω) and an even-complete subtree F n T n (n < ω) such that

(4.4) cl (G ) G (n < ω); P (n) n+1 n (4.5) G P = (n < ω); n ∩ (n) ∅ (4.6) G An is (n)-meager (n < ω); n ∩ P (4.7) if (n) F 1(n) T(F 1(n)) and (n) is even, then 2 ∈ \ | 2 | G A1(n) is (n)-meager (n, i < ω); n +F 1(n) _ P ∩ 2(n) i

1(n) 1(n) (4.8) if 2(n) T(F ) then Gn X A (n < ω). ∈ \ 2(n)

Since the topology < is ner than (n) (n < ω), cl (G ) G implies that P P P (n) n+1 n

cl < (Gn+1) Gn (n < ω), so we have that Gn = by (4.4) and Gn P by P n<ω 6 ∅ n<ω (4.5). By Lemma 4.17, (4.8) gives T T

G P An. n \ n<ω n<ω \ [ In the following construction we will successively grow the trees F n, so a node can be terminal after some intermediate step but not in the nal tree. To be concrete, we will

4.2. INTERSECTION CRITERIA 51

grow F n at in the nth step or never, while if s F n for s T n with (m) = (n, s) ∅ ∈ ∈ \ {∅} then we will grow F n at s in the mth step or never. We will declare when a tree does not grow any more from a node, such that this node remains terminal.

For n = 0, by (4.2) and (4.3) we have (0) = (0, ). To nd our G observe that X A0 ∅ 0 \ 0 is a () set containing P(0). By Corollary 4.14, P(0), (0) satises the conditions of 0 { } Theorem 4.7, in particular it is a 0 () topological Hurewicz test pair. By Theorem 4.7.1, 0 P and hence X A0 is (0)<-residual, by Corollary 4.8.2, X A0 is of (0)-second (0) \ P \ P 0 category, that is A is P (0)-meager in some nonempty basic P (0)-open set G.

If A0 is 0 () with 3 ϑ < ω then we have A0 = X A0 , so for some k < ω ϑ 1 i<ω \ (i) 0 0 and nonempty basic P (0)-open set G0 G we have thatT X A is P (0)-meager in G0. \ (k0) Put k_i F 0 (i < ω). Then (4.7) holds for n = 0 and (4.8) does not apply. Since P 0 ∈ (0) is (0)-nowhere dense we can pass to some basic (0)-open subset G G0 such that P P 0 G P = ; so (4.5-4.8) are satised for n = 0. 0 ∩ (0) ∅ 0 0 0 0 Else we have that A is 1() or 2(), that is A is P (0)-nowhere dense in G. In this case choose G G (P A0), put F 0 and F 0 does not grow any more, that 0 \ (0) ∪ ∅ ∈ is (0) = T(F 0). Now (4.7) does not apply; so we again have (4.5-4.8) for n = 0. 2 ∅ ∈

Suppose that Gn (n < N) is already dened such that (4.5-4.8) hold for n < N and

N 0 (4.4) holds for n < N 1; we nd our GN . Again, X A is a () set containing \ 0 P . By Corollary 4.14, P , (N) satises the conditions of Theorem 4.7, it is a (N) { (N) } 0 < < () topological Hurewicz test pair, P (N) = P (N 1), so GN 1 is P (N) -open. 0 N < By Theorem 4.7.1, P(N) and hence X A is P (N) -residual in GN 1, so by Corollary \ N N 4.8.2, X A is of P (N)-second category in GN 1, that is A is P (N)-meager in some \ nonempty basic P (N)-open set G GN 1. If AN is not a 0() or a 0() set with presentation then we have AN = X AN . 1 2 i<ω \ (i) N So for some kN < ω and nonempty basic P (N)-open set G0 G we have thatT X A is \ (kN )

52 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

N 0 0 P (N)-meager on G0. If A is a 1() or a 2() set with presentation then set G0 = G.

Since P(N) is P (N)-nowhere dense, we can pass to some basic P (N)-open subset

_ N G00 G0 such that G00 P(N) = and clP (N)(G00) GN 1. We put kN i F , (i < ω), ∩ ∅ ∈ then (4.4-4.6) hold for every basic (N)-open set G G00. P N

1(N) If (N) / T(F ) then set G = G00. If (N) is odd then neither (4.7) nor (4.8) 2 ∈ N | 2 | apply so the inductive step is complete. If (N) is even then (4.8) does not apply so | 2 | it remains to show (4.7). If (N) = then after the (N)th step of the construction 2 ∅ 1 1(N) 0 0 we had that A is neither a 1() nor a 2() set with presentation, 2 < and

1(N) G (N) A (N) is P (1(N))-meager (i < ω). By (4.3) we have that 1(N) < N. 1 +F 1 _i ∩ ∅ 1(N) 0 1(N) Since A (N) is () (i < ω), Lemma 4.15 gives that A (N) is P (N)-meager +F 1 _i 00 +F 1 _i ∅ ∅

in G1(N) hence also in GN , as required.

If (N) / T(F 1(N)), (N) is even but (N) = then we show that (N) will 2 ∈ | 2 | 2 6 ∅ 2 1(N) 1(N) never be a node of F . Let u be the terminal node of F on the branch of 2(N)

1(N) in T . By (4.2) there is an m < N such that 1(m) = 1(N) and 2(m) = u. After the

mth step of the construction u remained a terminal node of F 1(N), that is F 1(N) never

1(N) grows from u so 2(N) will never be a node of F . So again neither (4.7) nor (4.8) apply and the inductive step is complete.

If (N) T(F 1(N)) then we do the following. Let m < ω be such that (m) = (N) 2 ∈ 1 1 1(N) and 2(m) = 2(N) 2(N) 2. By (4.2) we have m < N. Since 2(N) = , A is neither || | 6 ∅ a 0() nor a 0() set with presentation and 2 < . Also, A1(N) is a 0 () set. We had 1 2 2(N) 00 th 1(N) (4.7) in the m step of the construction so by Lemma 4.15, G00 Gm implies that A 2(N) 1(N) 0 0 is (N)-meager in G00. If A is a () or a () set then it is actually (N)- P 2(N) 1 2 P

1(N) nowhere dense in G00 so we can nd a nonempty basic P (N)-open set GN G00 A . \ 2(N) 1(N) We do not grow F from the node 2(N), so (4.7) does not apply and (4.8) holds.

1(N) 0 1(N) 1(N) If A is () for some 3 ϑ < 0 then since A = X A _ , for 2(N) ϑ 2(N) i<ω \ 2(N) i T

4.3. A MOMENT OF BEING CONCRETE 53

1(N) some lN < ω and nonempty basic P (N)-open set GN G00 we have that X A _ \ 2(N) lN is (N)-meager on G . We put (N)_l_i F 1(N), (i < ω), then (4.7) holds and (4.8) P N 2 N ∈ does not apply. This completes the recursive step and the proof.

4.3 A moment of being concrete

0 In order to proceed we need to show at least one concrete () topological Hurewicz test pair. We do this as was done in [15].

Denition 4.19. We set (C , ) = (C, ), P = x C : m ω (x(m) = 1) , 1 C1 C 1 { ∈ 1 ∀ ∈ } 1 0 T1 = , P = P1 and r1 = ω. Suppose that the spaces (Cϑ, Cϑ ), the ϑ(Cϑ ) sets with {∅} ∅ ϑ presentation [Pϑ, (Pt )t Tϑ ] and the ordinal rϑ are dened for every ϑ < . Then let ∈

C = C , = , ϑi C Cϑi i<ω i<ω Y Y

(4.9) P = x C : i < ω (x(i, .) C P ) , { ∈ ∀ ∈ ϑi \ ϑi }

(4.10) T = n_t: t T , n < ω , { ∈ ϑn }

ϑn (4.11) P _ = C P C (t T , n < ω) n t ϑi t ϑi ∈ ϑn i

r = rϑi . i<ω X Claim 4.20. Let 0 < < ω and ϑ . We have C = 2r . The 0() set with pre- 1 i → 0 sentation [P, (Pt )t T ] satises {; in particular P, P is a () topological Hurewicz ∈ { }

54 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

< < test pair. We have = P and P = [ U,n : n < ω ] (1 < < ω1) where P i<ω ϑi P { } Q (4.12) U = (C P ) P C ,n ϑi \ ϑi ϑn ϑi i

single point so it is nowhere dense, as stated. Let now 1 < < ω1 and suppose that the statements are true for ϑ < . Then

r r r ϑi i<ω ϑi C = Cϑi = 2 = 2 = 2 . i<ω i<ω P Y Y By denition,

< P = = Pϑ , P(i) i i<ω i<ω _ Y as stated.

< Let now (C, G) (C, C , P). If C = X P and G is nonempty then G ∈ C \ ∈ P is nontrivial only on nitely many coordinates so it intersects X P = P . If \ i<ω (i) th (C, G) ϑi (C, j<ω,j=i P , P(i)) then by Claim 4.11, C is nontrivial onlyS on the i ∈ C 6 (j)

coordinate and GW= j<ω Gj where Gj = Cϑj except for nitely many j’s, Gj is basic

P(j) -open (j ω iQ) while Gi is basic Cϑ -open. Since [Pϑi , (Pt)t Tϑ ] satises {ϑi by ∈ \ { } i ∈ i the induction hypothesis, we have Pr (C) Pr (G) = , which implies C G = . So i ∩ i 6 ∅ ∩ 6 ∅ [P, (Pt )t T ] indeed satises . ∈ { < Finally we have P (C P ) = U,n, so by denition P = [ U,n : n < ω ]. (n) ∩ i

We point out a property of this construction.

< < Lemma 4.21. Let G C be a nonempty basic P -open set. Then int (cl (G)) = . P P 6 ∅

4.3. A MOMENT OF BEING CONCRETE 55

< Proof. If G is basic P -open then by regularity we have int < (cl < (G)) = G = . So P P 6 ∅ < let G be proper basic P -open, say G = G0 U,n where G0 is -open and n < ω. Since ∩ P < U,n is -dense in Cϑ Pϑ Cϑ , we have P i

< < G0 Cϑi Pϑn Cϑi int (cl (G)) ∩ P P i

so int < (cl < (G)) = , as stated. P P3 6 ∅ We close this section with the proof of Theorem 1.6.

Proof of Theorem 1.6. First we prove 1. and 2. in the = 2 case while this is exceptional in Theorem 1.2. Then we show 1. and 2. for 3 < ω and nally we treat 1 the extension for every 2 < ω . 1 So let = 2. We set

ω U 0 = x 2 : n N (x(n) = 0) (N < ω), 2,N { ∈ ∀ }

2 ω T 0 = (i): i < ω , P 0 = U 0 (i < ω), P 0 = 2 U 0 . 2 { } (i) 2,i 2 \ 2,N N<ω ! [ U Then the topology P20 is the renement of C1 by turning each point of the nite sets 20,N

(N < ω) into an open set. Clearly, P20 is the complement of a dense countable subset in ω 0 (2 , C1 ), so in particular P20 is 2(C1 ) and C1 -residual. As a complement of the dense U P P20 -open set N<ω 20,N , we also have that 20 is P20 -nowhere dense. Let A XS be 0() and take a continuous one-to-one mapping ϕ: (2ω, ) (X, ) 2 C1 → 1 1 ω such that ϕ (A) P 0 is of P P -second category in P 0. Then ϕ (A) (2 , C ) is ∩ 2 20 | 20 2 1 0 1 1 (C ) and ϕ (A) P 0 is of P P -second category; thus ϕ (A) is of C -second category, 2 1 ∩ 2 20 | 20 1 0 ω as well. Since a 2(C1 ) set in (2 , C1 ) is of second category only if its interior is nonempty, 1 1 ϕ (A) contains a nonempty -open set so ϕ (A) U 0 = for some N < ω. Then C1 ∩ 2,N 6 ∅ ϕ 1 A ( ), having nonempty interior, is of P20 -second category.

56 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

If A is not 0(), we apply Theorem 1.2 for A = A, A = X A. These sets 2 0 1 \ 0 cannot be separated by a ()-set, so since P20 is the complement of a countable dense ω ω subset of (2 , ), there is a continuous one-to-one mapping ϕ: 2 X with ϕ(P 0) A, C1 → 2 ω 1 ϕ(2 P 0) X A. So as we have seen above, ϕ (A) = P 0 is indeed P -meager. \ 2 \ 2 20 We turn to the 3 case. The Polish space (C , ) is obviously homeomorphic C to (C , ) (see e.g. [4], Theorem 7.4 on page 35). We show that P , fullls the 1 C1 { P } requirements for every 3 < ω . 1 Let A X be 0() for some < ω and take a continuous one-to-one mapping 1 1 1 ϕ: C X such that ϕ (A) P is of -second category in P . Then ϕ (A) C → ∩ P |P 0 1 is ( ) and ϕ (A) P is -residual in G P for some basic open set G. So C ∩ P |P ∩ 1 according to Claim 4.6, ϕ (A) is of P -second category, as required. Suppose now that A is not 0(). We apply Theorem 1.2 for A = A, A = X A. 0 1 \ 0 0 0 These sets cannot be separated by a ()-set, so since P is (C ) but not (C ), there is a continuous one-to-one mapping ϕ: C X with ϕ(P ) A, ϕ(2ω P ) X A. → \ \ 1 So by Denition 4.2.3, ϕ (A) = P is indeed P -meager.

0 Finally suppose that for some cardinal < 2ℵ , in our model the the union of a

0 number of meager sets is meager in Polish spaces. Let Ai (i < ) be () and 1 A = A be Borel. Since ϕ (A) P is of -second category in P , by our i< i ∩ P |P 1 assumptionS ϕ (A ) P is also of second category in (P , ) for some i < . So by i ∩ P |P 1 1 the rst statement, ϕ (A ) ϕ (A) is of -second category. This nishes the proof. i P

4.4 Constructible coverings

In this section we give the extension of Theorem 3.10.

Theorem 4.22. Let 1 < < ω1 be such that = 0 + 1 where 0 is a successor. Let (X, ) be an uncountable Polish space and P X be a proper 0() set. Then there is a

4.4. CONSTRUCTIBLE COVERINGS 57

mapping : 0(P ) 0(P ) such that B (B) and S → P

P (Bi) = (B, Bi 0(P ), i < ω). \ 6 ∅ ∈ S i<ω [

Proof. Fix our . First we construct = for (X, ) = (C, C ) and P = P. For

0 0 every B (P) x a decomposition B = Bj where Bj is (C ) (j < ω). Since ∈ S j<ω 0 the class 0 ( ) has the separation propertSy (see e.g. [4], (22.16) Theorem; and if = 2 0 C 0 note that C2 is zero dimensional) we can take a sequence (n(B))n<ω (C ) such 0 that

(4.13) B (B) C P (n < ω). i n \ (i) i n i n [ [ Set

(B) = n(B). m<ω m n<ω \ [ It is clear that (B) is 0( ) and (4.13) implies B (B) P . It remains to show C that if Bi 0(P ) with its xed decomposition Bi = Bi (i < ω) then we can nd ∈ S j<ω j a point in P (Bi). S \ i<ω n i n We apply ClaimS 4.18 for A = i n n(B ) (n < ω). We obtain that P n<ω A = . \ 6 ∅ Since S S (Bi) (Bi) An, n i n<ω n<ω [ [ we have P (Bi) = , which completes the proof of the special case. \ i<ω 6 ∅ If (X, ),SP are arbitrary, by Theorem 1.2 we can take a continuous one-to-one map

1 0 ϕ: (C , ) (X, ) such that ϕ (P ) = P . For B (P ) let C → ∈ S

1 (B) = (P ϕ(P )) ϕ( (ϕ (B))). \ ∪

Since homeomorphism preserves the Borel class of sets this denition makes sense and

fullls the requirements.

58 CHAPTER 4. TOPOLOGICAL HUREWICZ TEST PAIRS

With a more careful approach one could prove Theorem 4.22 for = 0 + 1 where 0 is a limit ordinal. We think that this is the maximum one can do in ZFC, that is Theorem 4.22 is consistently false when is a limit ordinal. It seems that proving Theorem 4.22

0 for such a is equivalent with an armative answer to the question of A. Miller for generated ideals, the consistency of which is not established at the moment. We return to this problem in Chapter 8.

Chapter 5

Testing the dierence hierarchy

In this chapter we show that for 0 < 3 a proper D (0) set cannot be well approx- 0 imated by a D0 () set for 0 < . As usual in this paper, approximation is measured by Baire category in a suitable topology. The interesting feature of this is that such a

0 testing is possible even if D() is not closed under taking countable union; this is why the result we obtain is weaker than what we have got used to for the Borel hierarchy.

The reasons for restricting are twofold. The more important one is that argument which follows is not applicable for = = ω. The less important, aesthetic reason is that for 3 all the ideas which are new comparing to the previous chapters appear without serious technicalities. We will discuss possible generalizations after the proof of Claim 5.5 and in Chapter 8.

We start with the denition of the relevant spaces. From now on in this section let 0 < 3 and 0 < < ω . 1

Denition 5.1. Take sequences (C (), ) and P (), ( < ω ) of copies of C() { P()} 1 the Polish space (C , ) and the 0( ) topological Hurewicz test pair P , dened C C { P } in Denition 4.19. Let

59

60 CHAPTER 5. TESTING THE DIFFERENCE HIERARCHY

C, = C(), C, = C() (0 < < ω1). < < Y Y The dierence operator will act on

U () = C () (C () P ()) C () ( < < ω ). , \ 1 < << ! [ Y Y We set

V () = P () (C () P ()) C () ( < < ω ), , \ 1 < << Y Y

V,() = P(), < Y

W () = P () C () ( < ω ). , 1 < < Y Y Now for every 0 < < ω1 let

P = D (U ()) = V (): , is odd is even , , , < { , ↔ } [

Q = C P = V (): , is odd is odd , , \ , { , ↔ } [ and dene the topologies

< < < < = , () = P () ( < ω1), , P() , P() < < < Y Y Y

= < (0)[ W (): ]. , , { , }

0 It is clear that U,() is (C, ) ( < < ω1). Observe that P,, Q, (0 < < ω1) are -open sets and that the sets V () ( ) are pairwise disjoint. Notice that , , < < W,() = V,() and , = ,() (0 < < ω1). The testing theorem can be stated as follows.

61

Theorem 5.2. Let 0 < 3 and 0 < < ω be xed and consider a D (0( )) set 1 C, A C . Let G be < -open; and if = 1 suppose that W () G in addition. Then if , , 1, ∈ A Q is -residual in G Q then A P is of -second category in G P . ∩ , , ∩ , ∩ , , ∩ , We formulate a corollary of this theorem.

Corollary 5.3. Let 0 < 3 and 0 < < ω be xed. 1 1. If A C is a D (0( )) set and Q A then A P is of -second , C, , ∩ , , category.

2. If A C is a D (0( )) set and P A then A Q is of -second , C, , ∩ , , category.

Proof. The rst statement is the special G = C, case of Theorem 5.2, while the second follows from Theorem 5.2 applied to A = C A and G = C . , \ , Corollary 5.3 has already the feature of dichotomy we are looking for: it allows to derive from the information that a set A is simply structured the conclusion that either it does not contain Q or it is big in category in C Q , and vice versa. , , \ , The proof of Theorem 5.2 is based on Claim 5.5 stating that sets like W,() behave

as topological Hurewicz test sets with the topology ,. Before stating and proving it we need the usual lemma on the coincidence of topologies.

Lemma 5.4. Let 0 < < ω and be xed. Then 1 1. for = 1, if G is < -open and W () G then G V () = ( ); 1, 1, ∈ ∩ 1, 6 ∅

2. for = 2, 3, V () is a < ()-dense set ( ); , ,

3. for = 2, 3, W () is a < ()-dense 0( < ) set ( ); , , 2 ,

4. if G is basic < (0)-open and G V () = then G is basic < ()-open ( ); , ∩ , 6 ∅ ,

62 CHAPTER 5. TESTING THE DIFFERENCE HIERARCHY

< 5. if G is basic -open and G V () = then there is a basic ()-open set G0 , ∩ , 6 ∅ , such that G V () = G0 V (); ∩ , ∩ ,

6. the topologies and < () ( ) coincide; ,|V,() , |V,()

7. if G is basic < (0)-open and G W () = then G is basic < ()-open ( ); , ∩ , 6 ∅ ,

< 8. if G is basic -open and G W () = then there is a basic -open set G0 such , ∩ , 6 ∅ , that G0 W () = G W (); ∩ , ∩ ,

9. the topologies and < () ( ) coincide. ,|W,() , |W,()

Proof. Since < = , 1 is obvious. By Denition 4.19, both P () and C () P () 1, C1, \ are < -dense ( < ) sets, so 2 follows. By Theorem 4.7.1 and Claim 4.5.1, P () is a P() < -residual 0( < ()) set ( < ), so we have 3. P() 2 P For 4, let G be basic < (0)-open with G V () = . If for some < , Pr (G) , ∩ , 6 ∅ C is proper basic -open then Pr (G) P () = hence G V () = , which is not P() C ∩ ∅ ∩ , ∅ the case. So Pr (G) is < -open ( < ), thus 4 holds. C P() < For 5, let G be basic -open, say G = G0 W () where G0 is basic (0)- , ∩ , , open and , such that G V () = . Since V () W () ( ) and ∩ , 6 ∅ , , V () W () = ( < ), we have G V () = G0 V () = . Thus by 4, G0 , ∩ , ∅ ∩ , ∩ , 6 ∅ < is basic ,()-open so 5 holds. Since < () ( ) is ner than < () and coarser than , 5 immediately gives 6. , , ,

For 7, observe that W,() = V,(). Thus G V, = for some . ∩ 6 ∅ So by 4, G is basic < ()-open (S ), as required. , < For 8, let G be a basic -open set, say G = G0 W () where G0 is basic (0)-open , ∩ , , and , and suppose that G W () = . From W () W () ( ) we get ∩ , 6 ∅ , , < G0 W () = G W () = thus from 7 we have that G0 is -open, as stated. ∩ , ∩ , 6 ∅ ,

63

Since < () ( ) is ner than < and coarser than , 9 immediately follows , , , from 8, so the proof is complete.

Claim 5.5. Let 0 < < ω , (Y, ) be an arbitrary nonempty Polish space and G 1 C Y be a nonempty < -open set, such that if = 1 then G (W () Y ) = in , , ∩ 1, 6 ∅ 0 addition. If A C, Y is (C, ) and A (W,() Y ) is of (, ) W,() Y - ∩ | second category in G (W () Y ) then there is a nonempty basic < -open set ∩ , , G G such that A is -residual in G and G (W () Y ) = . 0 , 0 0 ∩ , 6 ∅

Proof. Before starting the proof, observe that for = 2, 3 by Lemma 5.4.3, G ∩ (W () Y ) = for every nonempty < -open set G. , 6 ∅ , For = 1 we have that < = and that A is -open so every basic 1, C1, C1, -open set G can be chosen which satises G G A and W () Pr (G ). C1, 0 0 ∩ 1, ∈ C1, 0 0 0 Let now = 2 or = 3. Since A is (C, ), there is a 1(C, ) set B A and a nonempty basic -open set G? G such that G? (W () Y ) = , ∩ , 6 ∅ ? and B is (, ) W () Y -residual in G (W,() Y ). By Lemma 5.4.8 there is a | , ∩ basic < -open set G for which G (W () Y ) = G? (W () Y ). Then by , 0 0 ∩ , ∩ , < Lemma 5.4.9, B is , W,() Y -residual in G0 (W,() Y ). We show that B | ∩ is -residual in G ; then by B A, G fullls the requirements. , 0 0 < < For = 2 we have 2, = 2,() = C2, . So by Lemma 5.4.9,

< (2, ) W2,() Y = 2,() W2,() Y = C2, W2,() Y . | | |

Hence B is C2, W2,() Y -residual in G0 (W2,() Y ). By Lemma 5.4.3 we have | ∩ that W () Y is -residual in X Y so the 0( ) set B contains G . 2, C2, 1 C2, 0 This proves the = 2 case. For = 3 suppose that B is not -residual in G . Since C B is 0( ), 3, 0 3, \ 2 C3, 0 there is some nonempty basic -open set G0 G and ( ) set F C B 3, 0 1 C3, 3, \

64 CHAPTER 5. TESTING THE DIFFERENCE HIERARCHY

such that F is 3, -residual in G0, that is G0 F which gives cl < (G0) F . We can 3, < assume that G0 is not (0)-open, so G0 = (W () Y ) G00 for some where G00 3, 3, ∩ < < is basic (0) -open; thus by Lemma 5.4.7, G00 is () -open. Then for some nite 3, 3, < set I , PrC ()(G00) is basic -open ( I ), PrC ()(G00) is basic P ()-open 3 P3() ∈ ∩ 3 3 ( I ) and Pr (G00) = C () ( I). By Lemma 5.4.3 we have ∈ \ C3() 3 ∈ \

(5.1) G00 cl < () (G0) cl < (G0). 3, 3,

By Lemma 4.21,

< < int (cl (PrC3()(G00)) = ( I). P3() P3() 6 ∅ ∈

So by the regularity of PrY (G00) in Y ,

int < (cl < (G00)) = 3, 3, 6 ∅

which by (5.1) gives

H = int < (cl < (G0)) = . 3, 3, 6 ∅ Since G is regular < -open, 0 3,

H int < (cl < (G0)) = G0, 3, 3,

< < and from cl (G0) F we have H F . Since B is 3, W3,() Y -residual in 3, | G (W () Y ) and by Lemma 5.4.3, W () is < -residual in C , B is < - 0 ∩ 3, 3, 3, 3, 3, residual in G . But H G B, a contradiction. 0 0 \ In our present approach the validity of this claim limits our capacities in testing the dierence hierarchy. It is fairly simple to see that the proof of the lemma can be repeated

0 for 3 < < ω, 0 < < ω1; it is an imitation of the proof of Theorem 4.7 for the (C, )

set W,. Also, if < ω then the statement of the lemma follows for every 0 < < ω1 from the fact that P , is a topological Hurewicz test pair, using Kuratowski-Ulam { P }

65

Theorem as in the proof of Claim 4.5.5. The handicap of this approach is that Claim 5.5 is false for = = ω. This failure is the main reason why we restrict our attention to 3. In the proof of Theorem 5.2 the product structure of Denition 5.1 must be exploited. So we prove it in the following more general form. When Y is a single point, we get back Theorem 5.2.

Theorem 5.6. Let 0 < 3 and 0 < < ω be xed. Let (Y, ) be an arbitrary 1 nonempty Polish space and consider a D (0( )) set A C Y . Let G C Y C, , , be < -open, and for = 1 suppose that G (W () Y ) = in addition. Then if , ∩ 1, 6 ∅ A (Q Y ) is -residual in G (Q Y ) then A (P Y ) is of -second ∩ , , ∩ , ∩ , , category in G (P Y ). ∩ ,

Proof. We prove the statement by induction on . Let rst = 1, then A is a 0( ). By Claim 4.20, Claim 4.10 and Claim 4.5.5, P Y, is a topological C,1 { P } Hurewicz test pair in C Y , so the statement follows from Claim 4.6. ,1 Suppose now that 1 < < ω1 and that the statement holds for 0 < . Let A = D ((A ) ) with 0( ) sets A ( < ) satisfying A A ( < ). We < C, have that W () Y Q Y is -open. By assumption for = 1 and by , , , Lemma 5.4.3 for = 2, 3, G W () Y = so A is of -second category in ∩ , 6 ∅ , G W () Y . Thus there is a minimal such that the parity of and are dierent ∩ , and for some basic -open set G? G, A of -second category in the , , ? < nonempty G (W () Y ). Then by Lemma 5.4.8 there is a -open set G0 such ∩ , , ? that G (W () Y ) = G0 (W () Y ) = . ∩ , ∩ , 6 ∅ < We apply Claim 5.5 for A and G0; we obtain that for some nonempty basic - , open set G G0, A is -residual in G , and G (W () Y ) = . So in 0 , 0 ∩ 1, 6 ∅ particular, A is -residual in G (V () Y ), which is nonempty by Lemma , 0 ∩ ,

66 CHAPTER 5. TESTING THE DIFFERENCE HIERARCHY

5.4.1 if = 1 and by Lemma 5.4.2 if = 2, 3. But the parity of and dier, so V () Y P Y . That is if A A is also of -second category in , , \ < , G (V () Y ) then we are done. S 0 ∩ , Suppose that this is not the case. Then there is a < and a basic -open set , G? G such that G? (V () Y ) = and A is -residual in G? (V () Y ), 0 0 0 ∩ , 6 ∅ , 0 ∩ , and the parity of and dier, that is parity of and coincide. By Lemma 5.4.5,

< ? there is a basic () -open set G0 such that G (V () Y ) = G0 (V () Y ). , 0 0 ∩ , 0 ∩ , < Since G is basic -open, we can and do assume that G0 G . Then by Lemma 0 , 0 0 < 5.4.6, A is () -residual in G0 (V () Y ). By passing to a subset if necessary, , 0 ∩ ,

we assume that PrC()(G0) is proper P ()-open, that is

(5.2) Pr (G0 ) C () P (). C() 0 \ Set ˜ = ,

Y˜ = C () Y, ˜ = P() < ! < ! Y Y ˜ and G = G0. With this setting, using (5.2),

(5.3) G˜ W (˜) Y˜ = G0 (V () Y ) = ∩ ,˜ 0 ∩ , 6 ∅ and by Lemma 5.4.9,

< (5.4) (,˜ ˜) G˜ W (˜) Y˜ = ,˜ ˜ G˜ W (˜) Y˜ = | ∩( ,˜ ) | ∩( ,˜ ) < < = ,˜ ˜ G V () Y = ,() G V () Y . | 0∩( , ) | 0∩( , ) ˜ We apply Claim 5.5 in C,˜ Y for A, which by (5.3) and (5.4) is (,˜ ˜) W (˜) Y˜ - | ,˜ residual in the nonempty G˜ W (˜) Y˜ . We get that for some nonempty basic ∩ ,˜ < ˜-open set G˜ G˜, A is ˜-residual in G˜ and G˜ W (˜) Y˜ = . In ,˜ 0 ,˜ 0 0 ∩ ,˜ 6 ∅ particular, G˜ V () Y˜ = ( ˜) by Lemma 5.4.1 if =1 and by Lemma 5.4.2 0 ∩ ,˜ 6 ∅ if = 2, 3, and A is ˜-residual in G˜ V () Y˜ ( ˜). ,˜ 0 ∩ ,˜

67

Now G˜ G is -open and V () Y˜ is also -open ( ˜). So A 0 , ,˜ , is -residual in G˜ V () Y˜ for every ˜ with parity dierent from the , 0 ∩ ,˜ parity of ˜. By (5.2) and G˜ G0 , 0 0

Pr (G˜ ) C () P (), C() 0 \

so we have G˜ (V () Y ) = G˜ V () Y˜ = ( ˜). Then by Lemma 5.4.6, 0 ∩ , 0 ∩ ,˜ 6 ∅

(5.5) ( ) = < (0) = , G˜0 V () Y , G˜0 V () Y | ∩( , ) | ∩( , ) = < (0) ˜ = ,˜ G˜0 V () Y | ∩( , ) = < (0) ˜ = ( ˜) ( ˜). ,˜ G˜0 V () Y˜ ,˜ G˜0 V () Y˜ | ∩( ,˜ ) | ∩( ,˜ ) We get that A is ˜-residual in G˜ V () Y˜ for every ˜ with parity ,˜ 0 ∩ ,˜ dierent from the parity of ˜. Since A is also ˜-residual in G˜ V () Y˜ ,˜ 0 ∩ ,˜ ( ˜), this is possible only if D (A ) is ˜-residual in G˜ V () Y˜ < ,˜ 0 ∩ ,˜ for every ˜ with parity dierentfrom theparity of ˜. Set = ,

Y = C () Y˜ , = ˜, P() < ! < ! Y Y

A = D (A ) and G = G˜ H = where H is nontrivial only on the th coordinate < 0 ∩ 6 ∅ and Pr (H) isproper basic -open, i.e. Pr (H) C () P (). Since G˜ is C() P() C() \ 0 basic < ˜-open, it is < () ˜-open and so G is basic < -open. As above, we ,˜ ,˜ , have

G V () Y = G V () Y˜ = ( ) 0 ∩ , 0 ∩ ,˜ 6 ∅

68 CHAPTER 5. TESTING THE DIFFERENCE HIERARCHY

so by Lemma 5.4.6,

< (5.6) , G V () Y = ,(0) G V () Y = | 0∩( , ) | 0∩( , ) < = ,˜(0) ˜ G V () Y = | 0∩( , ) < = ,˜(0) ˜ G V () Y˜ = (,˜ ˜) G V () Y˜ ( ). | 0∩( ,˜ ) | 0∩( ,˜ ) Since A = D (A ) is ˜-residual in G˜ V () Y˜ , we get that A is < ,˜ 0 ∩ ,˜ -residual in G V () Y for every with parity equal to the parity of , ∩ , . That is, A is -residual in G Q Y . So by the induction hypothesis A is , ∩ , of -second category in G P Y . Since A = D (A ) A, this means , ∩ , < that A is of -second category in P Y . We have , ,

V () Y = V () Y ( < ) , ,

so by Lemma 5.4.6,

< (5.7) , V,() Y = ,(0) V,() Y = | | < = ,(0) V,() Y = | < = ,(0) V,() Y = (, ) V,() Y ( < ). | | Since and have the same parity,

P Y = P C () Y = , , < Y V (): < , is odd is even P Y. { , ↔ } , [ So A is of -second category in P Y , which completes the proof. , , Possible further extensions of testing theorems will be discussed in Chapter 8.

Chapter 6

-convergent functions I

It is a fact of life that the class of continuous real functions is not closed under pointwise convergence: instead, we obtain a realization of the Baire-1 functions. On the other hand, it is an easy exercise that the pointwise limit of a sequence of continuous functions with

length ω1 is necessarily continuous. This problem and other properties of the pointwise convergence of transnite sequences of real functions has been rst considered by W. Sierpinski [20]. In particular, he studied which class of functions will be closed under such convergencies. Since most of the classes, for example the class of Baire- functions for 2, are not, T. Natkaniec [17] introduced a stronger notion of pointwise convergence. We recall the precise setting in the following denition.

Denition 6.1. Let be a cardinal, (X, ) be a Polish space, (Y, d) be a separable metric space and consider an ideal on . We say that a sequence of functions f : X Y ( < I → ) -converges to the function f : X Y , in notation f f, if I → →I < : f (x) = f(x) { 6 } ∈ I for every x X. ∈ 69

70 CHAPTER 6. -CONVERGENT FUNCTIONS I

Similarly, we write f d f if for every ε > 0 and x X we have →I ∈ < : d(f(x), f (x)) > ε . { } ∈ I

In case of the ordinary ω1 convergence, as used in [5] and [20], we have = ω1 and

ω = [ω ] , that is the ideal of countable subsets of ω . However, our motivating theorem, I 1 1 answering Problem 1 in [17] on page 490, is related to the particular case, when the ideal contains the nite subsets of ω , that is = [ω ]<ω. 1 I< 1 Theorem 6.2. Let (X, ) be a Polish space, (Y, d) be a separable metric space, and for a xed < ω consider a family f : X Y ( < ω ) of Baire- functions. If f : X Y 1 → 1 → is such that f d f, then f is Baire-. →I< We note here that the original question asked by T. Natkaniec referred to -con- I< d vergence. However, it is easy to see that <-convergence implies < -convergence, so the I I d result above is formally stronger than the required. The suciency of < -convergence was I pointed out to the author by Petr Holicky. As W. Sierpinski showed ([20], Theorem 1 on page 133 and Theorem 2 on page 137),

ω for the class of continuous and Baire-1 functions Theorem 6.2 holds also for = [ω ] I 1 instead of . On the other hand, it is independent for every 2 < ω whether there I< 1 ω is a [ω1] -convergent sequence of Baire- functions whose limit function is Borel but not

d ω Baire- (observe that the < -convergence implies the [ω1] -convergence). The rst part I of the following theorem has already been proved by W. Sierpinski ([20], Section 6, pages 139 and 140) and further discussed by P. Komjath ([5], Theorem 3, page 499). Its second part, related to Problem 3 in [17] on page 490, is a simple analogue of Theorem 6.2.

Theorem 6.3. Let (X, ) be a Polish space, (Y, d) be a separable metric space.

ω 1. (W. Sierpinski, P. Komjath) Assuming CH, there exists an [ω1] -convergent se- quence of real Baire-2 functions whose limit function is not Borel.

71

0 < 2. Let < 2ℵ be an innite cardinal with cf() > ω and set = [] . For a xed J < ω , consider a family f : X Y ( < ) of Baire- functions and a Borel 1 → function f : X Y . If f d f and in our model the union of meager sets is → →J meager in Polish spaces, then f is necessarily Baire-.

The assumption on the additivity of meager sets holds under MA() (see e.g. [3], Theorem 1.2 on page 505 or [16], Theorem on page 170). The convergence of transnite

sequences of Baire-2 functions of length ω2 has also been investigated by P. Komjath

0 (see [5], Theorem 4 and Theorem 5 on page 500). It is consistent (with 2ℵ = ω2 and

MA(ω1)) that every real function can be obtained as such a limit. It is also consistent, under more complicated assumptions, that the limit function is necessarily Baire-2. The case when the underlying space X is not necessarily Polish but merely metric has been considered in [2] and [19]. Transnite convergence of derivatives is examined in [1] while in [18] transnite convergence of Baire- functions if treated under various set theoretic assumptions. In order to establish the connection between function classes and sublevel sets we will use the following classical result (see e.g. [4], Chapter II, Theorem 24.3 on page 190).

Theorem 6.4. Let (X, ) be a Polish space, (Y, d) be a separable metric space. Then for

1 0 every 1 < ω , a function f : X Y is Baire- if and only if f (U) X is () 1 → +1 for every open set U Y . In the metric space (Y, d), the open ball centered at x Y with radius is denoted by ∈ Bd(x, ). After these preparations, Theorem 6.2 and Theorem 6.3.2 are simple corollaries of Theorem 1.6.

Proof of Theorem 6.2. By Theorem 1 on page 133 and Theorem 2 on page 137 of

[20], the statement holds for 1. So let 2 < ω1 be xed and suppose that f d f →I< for a family f : X Y ( < ω ) of Baire- functions. → 1

72 CHAPTER 6. -CONVERGENT FUNCTIONS I Suppose that f is not Baire-. As the pointwise limit of the functions f : < ω , { } f is clearly Borel, so by Theorem 6.4, there is an open ball B (x, ) Y such that d 1 0 f (Bd(x, )) is Borel but not +1(). Set

1 1 H(ε) = f (B (x, ε)), H (ε) = f (B (x, ε)) d d

0 for every < ω1 and 0 < ε < . Note that by Theorem 6.4, H(ε) is in +1() for every

< ω1 and 0 < ε < .

0 Since H(0) is not +1(), by Theorem 1.6.2 there is a continuous one-to-one map ϕ: (2ω, ) (X, ) such that C1 →

(i) ϕ(P ) H(0), and +1

1 ω (ii) ϕ (H(0)) 2 is of rst category in the topology . P+1

1 By (i), there is an ε > 0 such that ϕ (H(ε )) P is of -second category. 0 0 ∩ +1 P+1 |P+1 1 Let J1(ε) denote that set of those indices < ω1 for which ϕ (H(ε)) is of P+1 -second category. We prove that ω J (ε) is nite for every ε < ε . Suppose that this is not true and 1 \ 1 0 d take a countably innite set J 0(ε) ω1 J1(ε). By the denition of < convergence, \ I ε < ε0 implies that

H(ε ) H0(ε) := H (ε), 0 J (ε) ∈[0 1 so we have that ϕ (H0(ε)) P is of -second category in P ; that is, since ∩ +1 P+1 |P+1 +1 0 1 H0(ε) is +1(), by Theorem 1.6.1 ϕ (H0(ε)) is of P+1 -second category. This is a 1 contradiction, since by the denition of J1(ε), ϕ (H0(ε)) is P+1 -meager.

ω So J1(ε) is of cardinality ω1 for every ε < ε0. In particular, given that (2 , P+1 ) has countable base, there is a -open set U 2ω such that for a countably innite set of P+1 1 indices J 00 J (ε /2) we have that ϕ (H (ε /2)) is -residual in U whenever J 00. 1 0 0 P+1 ∈

73

Hence for

H00 = H(ε0/2), J \∈ 00 1 ϕ (H00) is also -residual in U, so by (ii) we can nd a point x H00 H(0). Thus P+1 0 ∈ \ d f ( < ω1) is not < -convergent since I

ε0 J 00 < ω : d(f(x ), f (x )) > 1 0 0 2 n o is innite; a contradiction. The proof is complete.

Proof of Theorem 6.3.2. Again, for 1 the statement follows from the proofs of Theorem 1 on page 133 and Theorem 2 on page 137 of [20] ; so let 2. Now f is Borel by assumption; and the proof is the same as for Theorem 6.2, until the denition of J1. Now we show that card ( J (ε)) < for every ε < ε . \ 1 0 Suppose that this is not true and take a set J 0(ε) J (ε) of cardinality . By the \ 1 d denition of convergence, ε < ε0 implies that J

H(ε ) H0(ε) := H (ε), 0 J (ε) ∈[0

1 so we have that ϕ (H0(ε)) P is of -second category in P ; that is, by the ∩ +1 P+1 |P+1 +1 0 extension of Theorem 1.6.1, since H0(ε) is the union of the number of +1() sets H(ε) 1 ( J 0(ε)), ϕ (H0(ε)) is of -second category. Now this contradicts the assumption ∈ P+1 ω that the union of a number of meager sets is meager in (2 , P+1 ), since by the denition

1 of J (ε), ϕ (H (ε)) ( J 0(ε)) is -meager. 1 ∈ P+1

We continue as above; J1(ε) is of cardinality for every ε < ε0. In particular, given that cf() > ω and (2ω, ) has countable base, there is a -open set U 2ω such P+1 P+1 1 that for a set of indices J 00 J (ε /2) of cardinality we have that ϕ (H (ε /2)) is 1 0 0 -residual in U whenever J 00. Since in our model the intersection of a number P+1 ∈

74 CHAPTER 6. -CONVERGENT FUNCTIONS I

of P+1 -residual sets is again P+1 -residual, for

H00 = H(ε0/2), J \∈ 00 1 ϕ (H00) is also -residual in U, so by (ii) we can nd a point x H00 H(0). Again, P+1 0 ∈ \ this contradicts the d -convergence. The proof is complete. J

Chapter 7

Generalized separation and reduction

It is well known that a pair of disjoint analytic sets can be separated by a Borel set. If we need to estimate the Borel class of this separating set then Theorem 1.2 is very useful: we only need to test the pair of analytic sets via injections of 2ω. Since analytic sets also

have the generalized separation property, that is for every sequence (Ai)i<ω of analytic sets with A = there is a sequence (B ) of Borel sets such that A B (i < ω) i<ω i ∅ i i<ω i i and TB = , one can request a test for the complexity of the B ’s. We cannot refuse i<ω i ∅ i suchTa demand so we present test sets corresponding to this problem. If topologization seems to be articial for Theorem 1.2, here the topological feature of our Hurewicz test sets will be essential, namely for the proof that testing generalized separation is possible; this application is the main motivation of this section. The concepts are similar that of [7]. On the other hand we do not follow the spirit of [7], that is we do not dene closed games, neither will be reducing maps be one-to-one. We will have to refer to Borel determinacy, so in particular this proof is not eective. We also restrict our attention to sequences of Borel sets; all these simplication to make the

75

76 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

argument as short as possible for having a nice application of topological test sets. First we examine the problem of generalized separation; generalized reduction will follow by taking complements.

7.1 Witnessing generalized separation

We aim to prove the following theorem.

Theorem 7.1. For every 0 < < ω1, there exists a Polish space (D, D ) homeomorphic to (C, ) and 0 ( ) sets R (n) D (n < ω) such that the following hold. C +1 D 1. If B D is a 0( ) set and R (i) B (i < ω) then B = . i D i i<ω i 6 ∅ T 2. If (X, ) is a Polish space and (Ai)i<ω is a sequence of Borel sets in X such that A = then i<ω i ∅ T (a) either there is a sequence (B ) of 0() sets such that A B (i < ω) and i i<ω i i B = , i<ω i ∅ (b) orT there is a continuous map ϕ: D X such that ϕ(R (i)) A (i < ω). → i

Observe that if 2b holds then the map ϕ indeed shows that 2a fails: if (Bi)i<ω a

0 1 sequence of () sets such that A B X (i < ω) then ϕ (B ) = follows i i i<ω i 6 ∅ 1 from 1 using R (i) ϕ (B ) (i < ω), so B = . So thisT theorem is of Hurewicz i i<ω i 6 ∅ type. T Before starting the construction we feel obliged to clarify why the testing is set up

0 0 only for sets and not for sets. We have three reasons for this, the rst is a corollary of Theorem 7.1.

Corollary 7.2. If in Theorem 7.1 case 2a holds then the sequence (Bi)i<ω can be chosen such that B 0() (i < ω). i ∈

7.1. WITNESSING GENERALIZED SEPARATION 77

0 Proof. Since the class has the generalized separation property in Polish spaces (see e.g. [4], (22.16) Theorem on page 172), the statement follows.

0 That is generalized reduction with sets is stronger than generalized reduction with 0 sets. The following result says that in a particular case Theorem 1.2 gives an answer 0 to the problem for sets .

Claim 7.3. Let A , A X be disjoint analytic sets. Then there are disjoint 0() sets 0 1 B , B X such that A B (i = 0, 1) if and only if A can be separated from A by a 0 1 i i 0 1 0 0 () set and vice versa, A1 can be separated from A0 by a () set.

Proof. If we have the disjoint 0() sets B , B X such that A B (i = 0, 1) 0 1 i i 0 then the separation is obviously possible in both directions. Suppose now that a () 0 set Bi0 separates Ai from A1 i (i = 0, 1). The class has the reduction property in 0 Polish spaces (see again [4], (22.16) Theorem on page 172), thus we have disjoint ()

sets Bi Bi0 (i = 0, 1) such that B0 B1 = B0 B10 . Since Bi0 A1 i = (i = 0, 1), we ∪ ∪ ∩ ∅

have Bi A1 i = (i = 0, 1). But A0 A1 B0 B10 = B0 B1, so we get Ai Bi ∩ ∅ ∪ ∪ ∪ (i = 0, 1). The proof is complete.

The next clam is simply trivial.

Claim 7.4. Let 0 < < ω be xed and let ϑ . Let (A ) be a decreasing sequence 1 i → i i<ω of analytic sets in X such that A = . If there is a sequence (B ) of 0() sets i<ω i ∅ i i<ω such that A B (i < ω) andT B = then there is a sequence (B0) satisfying i i i<ω i ∅ i i<ω 0 Ai B0 (i < ω) and B0 = Tsuch that B0 is () (i < ω). i i<ω i ∅ i ϑi

T 0 Proof. Write Bi = Bi(j) where Bi(j) is () (j < ω) and Bi(j) Bi(k) (k j<ω ϑj 0 j < ω, i < ω). Set Bi0 =T j i Bj(i), then Bi0 is ϑi () (i < ω). Since Ai Aj Bj Bj(i) (j i < ω) we have A TB0 (i < ω). We also have B0 = B (i) = B = , i i i<ω i i,j<ω j i<ω i ∅ which completes the proof. T T T

78 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

Claim 7.3 and Claim 7.4 handle the two extreme situations in generalized separation: when we have only two sets which are disjoint or we have innitely many sets which are decreasing. The cases in-between can be handled with similar tricks but the sepa- ration conditions become numerous and complicated, so we omit them and turn to the

0 construction of test sets for generalized separation by sets.

Denition 7.5. For = 1 let (D , ) = (ω + 1, o ). Let : ω ω ω be the 1 D1 ω+1 1 → bijection dened by

1 (n, i) = (i + n + 1)(i + n) + n (n, i < ω) 1 2

and set R (n) = (n, i): i < ω (n < ω) and R = ω . 1 { 1 } 1 { } For 1 < < ω , let (C (i), ), P (i), (i < ω) be copies of the space (C , ) 1 C(i) { P(i)} C and 0 Hurewicz test pair P , dened in Denition 4.19. Set { P }

D = C(i), D = C(i), i<ω i<ω Y Y Pˆ (n) = C (i) P (n) C (i), i

< < ˆ = P (n) P(n) P(i) P(i) i

R (n) = D Pˆ (n) Pˆ (m) = P (i) (C (n) P (n)) P (n), \ ∩ \ m<ω,m=n i

R = Pˆ (i) = P (i), < = < , R P(i) i<ω i<ω i<ω \ Y Y .

We state a trivial lemma on the Borel class of the preceding sets.

7.1. WITNESSING GENERALIZED SEPARATION 79

ˆ 0 0 Lemma 7.6. For every 1 < < ω1, P(n) is a (D ) set and R(n) is a +1(D ) set (n < ω).

Next we show that the sets R(n) do the job for Theorem 7.1.

Claim 7.7. Let 0 < < ω be xed and suppose that B D is a 0( ) set and 1 i D R (i) B (i < ω). Then R B = . i ∩ i<ω i 6 ∅ T 0 Proof. Consider rst = 1. If Bi is (D ) then ω cl (R1(i)) Bi (i < ω) so 1 1 ∈ D1 indeed ω R B . ∈ 1 ∩ i<ω i Let now 1 < T< ω . By Claim 4.5.5, Pˆ (n), (n < ω) is a topological Hurewicz 1 Pˆ(n) test pair in the Polish space n o

< D, C (n) P <(i) . P(i) i

Moreover, also by Claim 4.5.5, Theorem 4.7 holds for Pˆ(n) with the presentation inherited

< from the presentation of P dened in (4.9-4.11) of Denition 4.19. We also have = Pˆ(n) < (n < ω). So by Theorem 4.7.1, Pˆ (i) (i < ω) hence R is < residual in D . We show R R that B (i < ω) is also < -residual in D , this will complete the proof. i R

Fix some i < ω. Since D Pˆ(i) is ˆ -residual and Pˆ(j) is also ˆ -residual \ P(i) P(i) (j < ω, j = i), R(i) and thus Bi are also ˆ -residual in D (i < ω). Since Bi is 6 P(i) 0 (D ) , it is also

0 < < C(n) P (i) . P(i) i

in D, as stated.

The next task is to nd the appropriate image of R(n) (n < ω) if generalized separa- tion by 0 sets is not possible. Fix a 1 < < ω . Let : ω r ω be a bijection; using 1 →

80 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

ω D = (2r ) by Claim 4.20, let : C D denote the corresponding homeomorphism. → We dene the Borel game G, which is the heart of the proof of Theorem 7.1. Let A , G, H D be Borel sets such that A G (i < ω). In the game G ((A ) , G, H) i i i i<ω players I and II play

I (0) (1) . . . II (0) (1)

where (i), (i) 0, 1 (i < ω), and II wins if and only if ∈ { }

() R (i) = () A (i < ω), ∈ ⇒ ∈ i () G and ∈ * () R = () / H; ∈ ⇒ ∈ else I wins.

We associate maps to the strategies of I.

Denition 7.8. Fix a 1 < < ω and let be a strategy of I in G . Then : D 1 → D denotes the function mapping to a y D the unique x D such that the run ∈ ∈ 1 1 (x), (y) is according to . { }

Note that is continuous. In the sequel we will use this property in the form that

1 keeps the Borel class of sets.

By the following claim the determinacy of the game G indeed proves Theorem 7.1.

Claim 7.9. For every 1 < < ω1, in the game G a winning strategy

1. for player I gives a sequence (B ) of 0( ) sets such that A B (i < ω) and i i<ω D i i (G H) B = ; \ ∩ i<ω i ∅ T

7.1. WITNESSING GENERALIZED SEPARATION 81

2. for player II gives a continuous map ϕ: D G such that ϕ(R ) H = and → ∩ ∅ ϕ(R (i)) A (i < ω). i

1 ˆ Proof. Consider rst a winning strategy for I. Let Bi = (P(i)) (i < ω). By Lemma 7.6, B is 0( ) (i < ω). So it remains to show that A B and that i D i i (G H) B = . \ ∩ i<ω i ∅ 1 SupposeTthat for some i < ω we have a y A B , that is y A (Pˆ (i)). Since ∈ i \ i ∈ i \ 1 1 A (Pˆ (i)) = A (D Pˆ (i)), there is some x D Pˆ (i) such that the run i \ i ∩ \ ∈ \ 1 1 (x), (y) in G is according to . We have D Pˆ (i) = R (i) D Pˆ (j) Pˆ (i) . \ ∪ \ ∪ j<ω,j=i [6 But if x R (i) then II wins since y A G while if x / R (i) and ∈ ∈ i ∈

x D Pˆ (j) Pˆ (i) ∈ \ ∪ j<ω,j=i [6 then x / R (j) (j < ω) and x / R so II wins already by y G. This contradicts the ∈ ∈ ∈ denition of and proves that A B (i < ω). i i Suppose now that y (G H) B , that is y G H and ∈ \ ∩ i<ω i ∈ \ T 1 (7.1) y (Pˆ (i)) (i < ω). ∈

1 1 Let x D be arbitrary such that the run (x), (y) in G is according to . Then ∈ by (7.1), x Pˆ (i) so x / R (j) (j < ω). Thus II wins by y G H, which again ∈ i<ω ∈ ∈ \ contradicts theTchoice of and proves (G H) B = . \ ∩ i<ω i ∅ Let now be a winning strategy of II. Let ϕ: TD D map to an x D the unique → ∈ 1 1 y D such that the run (x), (y) in G is according to . This map is clearly ∈ continuous. We have ϕ(D ) G, ϕ(R ) H = and if i < ω and x R (i) then ∩ ∅ ∈ ϕ(x) A since the strategy is winning for II. This completes the proof. ∈ i

82 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

From this Theorem 7.1 will follow by standard arguments (see [7], Corollary 2 and Theorem 3); that we give in the remaining part of this section. We have to extend the results to arbitrary Polish spaces; for this we need the following result (see e.g. [7], Theorem on page 455).

Theorem 7.10. (Saint Raymond) Let (E, E), (F, F ) be compact metrizable spaces and (Q, ) be a Polish space. If h: E F is a continuous surjection and f : E Q is a rst Q → → class function then h has a rst class section (i.e. there is a Baire-1 function s: F E → such that h(s(y)) = y (y F )) such that f s: F Q is also a rst class function. ∈ →

0 This theorem is applicable for a class of sets if this can be built up from 2 sets on some canonical way.

Denition 7.11. Let D 2ω and let (C(n)) be a sequence of arbitrary subsets of n<ω some base set X. The the Hausdor operation associated to D acts on the sequence

(C(n))n<ω as D((C(n)) ) = x X : n < ω : x C(n) D . n<ω { ∈ { ∈ } ∈ } A class is called a 0 generated Hausdor class if there is a basis D 2ω such that in 2 every Polish space (X, ) the members of are the sets of the form D((C(n))n<ω) where (C(n)) 0(). n<ω 2

0 0 We recall that each class (1 < < ω1) is a 2 generated Hausdor class (see e.g. [4], (23.5) Exercise on page 180.) The following corollary is a variant of the so-called transfer lemma (for a version, see e.g. [7], Corollary 2 on page 455). It will allow us to extend our results to arbitrary compact spaces.

Corollary 7.12. Let (E, ), (F, ) be compact metrizable spaces, h: E F be a con- E F → 0 tinuous surjection and be a 2 generated Hausdor class. Suppose that (Ai)i<ω is a

7.1. WITNESSING GENERALIZED SEPARATION 83

sequence of subsets of F , T F and there is a sequence (H ) of subsets of E i i<ω 1 1 satisfying h (A ) H = (i < ω) and H = h (T ). Then there is a sequence i ∩ i ∅ i<ω i (B ) of subsets of F satisfying A SB = (i < ω) and B = T . i i<ω i ∩ i ∅ i<ω i S 0 Proof. Let Hi = D((Ci(n))n<ω) where D is a basis for and Ci(n) is 2(E) (i, n < ω). Let f : E [0, 1] be a rst class function such that C (n) = x E : f (x) > 0 (i, n < i,n → i { ∈ i,n } ω ω ω). Set f : E [0, 1] , f = f . From Theorem 7.10 for h and f we get a rst → i,n<ω i,n class section s of h such that f Qs is also of rst class, that is V (n) = y F : f (s(y)) > i { ∈ i,n 0 is a 0( ) set. Thus for B = D((V (n)) ) we have B (i < ω). It is clear } 2 F i i n<ω i ∈ 1 1 1 1 that s (Hi) = Bi (i < ω). Since s (h (T )) = T and i<ω Hi = h (T ), we obtain 1 B = T . From s (H) = B we have s(B ) H , whicSh gives B = h(s(B )) h(H ) i<ω i i i i i i i i 1 (Si < ω). Since H h (A ) = implies h(H ) A = (i < ω), we have B A = i ∩ i ∅ i ∩ i ∅ i ∩ i ∅ (i < ω) which completes the proof.

It remains to nish with the proof of Theorem 7.1. We prove a slightly stronger extension; Theorem 7.1 is the A = case of Theorem 7.13 together with Claim 7.7. i<ω i ∅ T Theorem 7.13. Let 0 < < ω1. In the Polish space (X, ) let (Ai)i<ω be a sequence of Borel sets. Then

1. either there is a sequence (B ) of 0() set such that A B (i < ω) and i i<ω i i

i<ω Bi = i<ω Ai, T T 2. or there is a continuous map ϕ: D X such that ϕ(R ) A = and → ∩ i<ω i ∅ ϕ(R (i)) A (i < ω). T i

Proof. First we show that 2 excludes 1. If a map ϕ satises the conditions of 2

1 then by Claim 7.7, R ϕ (B ) = . This implies ϕ(R ) B = so since ∩ i<ω i 6 ∅ ∩ i<ω i 6 ∅ ϕ(R ) X A , we Thave B = A . T \ i<ω i i<ω i 6 i<ω i T T T

84 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

To return to the proof of the theorem, set H = i<ω Ai, and let rst = 1. If cl (A ) = H then we have 1. Else let x Tcl (A ) H. Since x cl (A ) i<ω i ω ∈ i<ω i \ ω ∈ i T(i < ω), we have T

A B (x , 1/n) = (i, n < ω) i ∩ ω 6 ∅

where B (xω, 1/n) is the open ball around xω with radius 1/n (n < ω). So we can pick points x A B (x , 1/n) (i, n < ω). Let ϕ: D X be dened by ϕ(i) = x 1(n,i) ∈ i ∩ ω 1 → i (i ω). This map is clearly continuous, ϕ(R (i)) A (i < ω) and ϕ(R ) / H, as 1 i 1 ∈ required.

Let now 1 < < ω1 and let (Ai)i<ω be a sequence of Borel sets in (X, ) such that ˆ i<ω Ai = H. Let (X, ˆ) be a Polish compactication of (X, ) (see e.g. [4], (4.14) TTheorem); then X Xˆ is a 0(ˆ) set. Take a continuous surjection h: D Xˆ and 2 → consider the game

1 1 1 G((h (Ai))i<ω, h (X), h (H)).

This game is clearly Borel so by Borel determinacy either player I or player II has a winning strategy.

If player I has a winning strategy then by Claim 7.9.1 we have a sequence (Ui)i<ω of

0 1 ( ) sets such that h (A ) U (i < ω) and D i i

1 1 (h (X) h (H)) U = . \ ∩ i ∅ i<ω \ That is by

1 1 1 h (H) h (A ) h (X) (i < ω) i we have

1 1 (7.2) h (X) U = h (H). ∩ i i<ω \

7.2. GENERALIZED REDUCTION 85

Let (E, ) = (D , ), (F, ) = (Xˆ, ˆ), = 0( ), T = Xˆ H and H = D E D F D \ i \ 1 1 1 1 (h (X) U ) (i < ω). Since h (A ) h (X) U implies h (A ) H = (i < ω) ∩ i i ∩ i i ∩ i ∅ and

1 1 1 H = D h (X) U = D h (H) = h (T ) i \ ∩ i \ i<ω i<ω ! [ \ 0 by (7.2), we can apply Corollary 7.12 to get a sequence (Vi)i<ω of (ˆ) sets such that A V = and V = Xˆ H. Then B = X V is 0(), A B (i < ω) and i ∩ i ∅ i<ω i \ i \ i i i

i<ω Bi = H, as required.S

1 T If Player II has a winning strategy then we have a continuous map : D h (X) → 1 1 such that (R ) h (H) = and (R (i)) h (A ) (i < ω). Then for ϕ = h , ∩ ∅ i ϕ: D X is also continuous and satises ϕ(R ) H = , ϕ(R (i)) A (i < ω). This → ∩ ∅ i completes proof.

7.2 Generalized reduction

Generalized reduction follows from Theorem 7.13 by taking complements.

Theorem 7.14. Let 0 < < ω1. In the Polish space (X, ) let (Ai)i<ω be a sequence of Borel sets. Then

1. either there is a sequence (B ) of 0() set such that B A , B B = i i<ω i i i ∩ j ∅ (i, j < ω, i = j) and B = A , 6 i<ω i i<ω i S S 2. or there is a continuous map ϕ: D X such that ϕ(R ) A and ϕ(R (i)) → i<ω i ∩ A = . (i < ω). S i ∅

Proof. By repeating the argument in the proof of Theorem 7.13 one easily obtains

that by Claim 7.7, 2 excludes 1. Set A0 = X A (i < ω). By Theorem 7.13, either there i \ i

86 CHAPTER 7. GENERALIZED SEPARATION AND REDUCTION

0 is a sequence (B0) of () set such that A0 B0 and A0 = B0 or there is a i i<ω i i i<ω i i<ω i continuous map ϕ: D X such that ϕ(R ) A0 =T and ϕ(RT(i)) A0 (i < ω) . → ∩ i<ω i ∅ i 0 In the rst case let B00 = X B0 (i < ω). TheT class has the generalized reduction i \ i 0 property so there is a () set B B00 (i < ω) such that B B = (i, j < ω, i = j) i i i ∩ j ∅ 6

and i<ω Bi = i<ω Bi0 = i<ω Ai; which is 1.

InSthe secondS case we Shave ϕ(R (i)) A = (i < ω) and ϕ(R ) X A0 = ∩ i ∅ \ i<ω i

i<ω Ai; which is 2. So the proof is complete. T

S 0 0 For reduction by sets the same remarks apply as for separation by sets (see Corollary 7.2, Claim 7.3, Claim 7.4 and their discussion).

Chapter 8

Concluding remarks

0 We constructed topological Hurewicz test pairs for every 0 < < ω1 and we proved the applications announced in the introduction. To conclude this thesis we would like to present some problems which were left open implicitly or explicitly in the preceding chapters.

Problem 8.1. Extend Theorem 4.7 or Theorem 5.2 to the entire Borel-Wadge hierarchy.

The aesthetic motivation for this is the fact that Theorem 1.2 has its extension to all the Borel-Wadge degrees (see [6] and [8]). The real motivation is that it is not clear at

all how one could get rid of the rigid structure of the sets P (0 < < ω1), which were practically our only examples of Hurewicz test sets. The reason why we did not prove Theorem 5.2 for 3 < was the failure of Claim 5.5 for = = ω. But the reason for this failure is that we cannot allow too much exibility in the structure of our topological Hurewicz test sets so the proofs get lost in technicalities. An approach which is able to handle the entire Borel-Wadge hierarchy will hopefully be free of these superuous complications. Next let us recall the theorem of S. Solecki mentioned in the introduction ([21], The-

87

88 CHAPTER 8. CONCLUDING REMARKS

orem 1 on page 1023).

Theorem 8.2. (S. Solecki) Let be a family of closed subsets of a Polis space (X, ). I Let A X be a 1() set. Then either A can be covered by a countable union of members 1 of or there is a 0() set G A such that F G relatively meager in G for every I 2 ∩ F . ∈ I

Less formally this result says that if a -ideal is generated by its 0 members I 1 the problem whether an analytic set belongs to or not depends only on sets of low I 0 complexity, namely on 2 sets. This led to the formulation of Question 1.5 and Question 3.16. It is important to note that these questions are already refuted by the following unpublished result of A. Kechris and M. Zeleny.

Theorem 8.3. (A. Kechris-M. Zeleny) Assume V = L. Then there is an analytic set A C and a -ideal strongly generated by its 0( ) members such that contains I 2 C I every Borel subset of A.

However, in view of Theorem 3.15 it seems to be interesting to look for an alternative counterexample. Mostly because in the proof of Theorem 3.15, CH was used for having

only ω1 many sets to handle, so this result is more “descriptive” than “set theory”.

Problem 8.4. Prove (without V = L) for every 2 < ω that the answer to Question 1 1.5 and/or to Question 3.16 is consistently negative.

We know that using the approach of nested sequences applied in the proof of Theorem 3.15 it can be proved that Question 1.5 has a negative answer under CH; this will be published elsewhere. The reason why we could not include it in this thesis is the same as above: the proof gets extremely technical due to the fact that the structural conditions

on P4 are so restrictive that this set cannot be used. We admit that we used PL instead

89

of P3 for the same reason. And we also admit that this approach seems to be completely inadequate to handle the 4 < ω case. It seems that Problem 8.4 is in close relation 1 also to the question whether Theorem 4.22 holds for = 0 + 1 where 0 is a limit ordinal.

Problem 8.5. Prove for every 2 < ω that the answer to Question 1.5 and/or to 1 Question 3.16 is consistently positive.

We have no results in this direction. It can easily happen that the conditions for a positive answer are much more natural and easier to handle. Since the proof of Theorem 8.2 is as elegant as numerous its corollaries are, it would be nice if an analogous statement was true for higher Borel classes, at least consistently.

Problem 8.6. Make the arguments in Chapter 7 eective.

It seems to require only endurance. But a lot more than eective generalized separation and reduction might be achieved. We could ask for a much closer analogy with the Baire Category Theorem and related techniques than what we have up till now through our

topological Hurewicz test pairs. Namely for 1 < < ω1, we intend to have some machinery

0 which isolates a “maximal” set inside a given Borel set which behaves like “taking the interior”. The best candidate for the framework of such a theory seems to be the eective

0 machinery. It may allow to dene topologies P to a much larger family of set in such a way that P, is a topological Hurewicz test pair. We may have an elegant, natural { P } proof for Theorem 4.22 and for ideal generation theorems in general. And we may be in the position to answer Question 1.5 and Question 3.16: in the negative by repeating the relatively simple argument of the proof of Theorem 3.15 and in the positive by repeating the absolute simple argument of the proof of Theorem 8.2 with a properly chosen notion of an “eective closure”.

90 CHAPTER 8. CONCLUDING REMARKS

Bibliography

[1] M. Dindos, Limits of Transnite Convergent Sequences of Derivatives, Real Anal. Exchange 22 no. 1 (1996/97), 338–345.

[2] M. Elekes, K. Kunen, Transnite sequences of continuous and Baire Class 1 func- tions, Proc. Amer. Math. Soc. 131 no. 8 (2003), 2453–2457.

[3] I. Juhasz , Consistency results in topology, Handbook of Mathematical Logic (1977), 503–522.

[4] A. S. Kechris, Classical Descriptive Set Theory, Graduate Texts in Mathematics 156, Springer-Verlag (1994).

[5] P. Komjath , Limits of transnite sequences of Baire-2 functions. Real Anal. Ex- change 24, no. 2, (1998/99), 497–502.

[6] A. Louveau, Some results in the Wadge hierarchy of Borel sets, Cabal seminar 79– 81, Lecture Notes in Math., 1019, Springer, Berlin, (1983), 28–55.

[7] A. Louveau, J. Saint Raymond, Borel Classes and Closed Games: Wadge-type and Hurewicz-type Results, Trans. Amer. Math. Soc., 304, No. 2, (1987), 431–467.

[8] A. Louveau, J. Saint-Raymond, The strength of Borel Wadge determinacy. Cabal Seminar 81–85, Lecture Notes in Math., 1333, Springer, Berlin, (1988), 1–30.

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[9] T. Matrai , The ω1-limit of Baire-2 functions in Baire-2, Ann. Univ. Sci. Budapest. Eotvos Sect. Math. 46, (2003), 167–174.

[10] T. Matrai , On the closure of Baire classes under transnite convergences, Fund. Math. 183, (2004), 157–168.

0 [11] T. Matrai , 2-generated ideals are unwitnessable, Fund. Math., submitted for pub- lication.

[12] T. Matrai , Hurewicz test sets for generalized separation and reduction, Math. Proc. Cambridge Phil. Soc., submitted for publication.

0 0 [13] T. Matrai , Covering -generated ideals by sets, J. London Math. Soc., sub- mitted for publication.

[14] T. Matrai , Topological aspects of Hurewicz tests for the dierence hierarchy, preprint.

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[20] W. Sierpinski , Sur les suites transnies convergentes de fonctions de Baire. Fund. Math. 1, (1920), 132–141.

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Index

d B , 32 < -convergence, 70 I

B , 32 , 80

C, 11 1(n, i), 78

C,, 59 , 79

C, 53 o, 11

cl , 13 Pt , 53

{, 46 P, 53 (X, , P ), 45 P , 23 C L Dϑ(.), 14 PL-nested, 32

0 Dϑ () , 14 P,, 60

D,78 Pˆ(n), 78

diam , 13 PrX , 13

f f, 69 , 39 →I P 0 f d f, 70 (P ), 14 →I P , 14 , 57

0 G, 80 (), 14 (ϑ), 38 Q , 60 HX,P , , 70 R , 78 I< int , 13 R(n), 78 -convergence, 7, 69 r , 53 I , 34 , 80 IPL 94

INDEX 95

s t, 11 (n), 47 P s t, 11 ϑi , 11 → _ s i, 11 U,n, 54

0(P ), 14 U,(), 60 S 0 V,(), 60 (), 14

< , 53 n<ω n, 12 P

< W ,(), 60 ,, 60 ZX,PL (A), 23 P , 53 ZX,PL (A), 23 , 78 s Pˆ(n) X,P ZG (A), 21 ,, 60

Ts, 11 basic open set, 13

T, 53 class t+T , 49 additive, 14 T(T ), 11 dual, 14 [], 32 multiplicative, 14 [ ], 12 P < [], 32 dierence hierarchy, 14 PL <, 18, 39 P even-complete subtree, 49 < , 78 R < generator of ideal, 15 ,(), 60 strong, 15 C, , 59

C , 53 Hausdor class

0 C , 11 2 generated, 82

PL [], 32 Hausdor operation, 82

P , 18, 39 Hurewicz test

96 INDEX

for F sets, 6 property, 15

0 for (2ω ) sets, 6 generalized, 15 Hurewicz, W., 5 Sierpinski, W., 70 Solecki, S., 7, 87 Kechris, A., 88 Komjath, P., 70, 71 topological Hurewicz test pair

0 Kuratowski-Ulam Theorem, 39 3(), 17 0(), 38 Louveau, A., 6 Lusin, 17, 22 Zeleny, M., 88

Miller, A., 7, 9, 58

Natkaniec, T., 7, 70 nested sequence, 32 nontrivial on a coordinate, 14

presentation

0 for 1() sets, 37 0 for () sets, 37 proper

0 Dϑ () set, 14 0 ( ) set,14 basic -open set, 13

reduction property, 15 generalized, 15

Saint Raymond, J., 6, 82 separation, 15

INDEX 97

Department of Mathematics and its Applications Central European University Nador u. 6. 1051 Budapest, Hungary e-mail: [email protected]